Let \(\vec{r}=\vec{a}+t \vec{b}\) be the equation of the given line which is not parallel to the given plane whose equation is \(\text { So, } \vec{b} \cdot \vec{n} \neq 0\) .
Let \(\vec{u}\) be the position vector of the meeting point of the line with the plane. Then \(\vec{u}\) satisfies both \(\vec{r}=\vec{a}+t \vec{b}\) and \(\vec{r} \cdot \vec{n}=p\) for some value of t , say t₁. So, We get
\( \vec{u}=\vec{a}+t \vec{b} \) ...(1)
\(\vec{u} \cdot \vec{n}=p\)   ...(2)
Substuting (1) in (2), we get
\( \left(\vec{a}+t_{1} \vec{b}\right) \cdot \vec{n}=p \)
\(\text { or } \quad \vec{a} \cdot \vec{n}+t_{1}(\vec{b} \cdot \vec{n})=p \ \)
\(\text { or } \ \ t_{1}=\frac{p-(\vec{a} \cdot \vec{n})}{\vec{b} \cdot \vec{n}}\)
\(\text { or } \ t_{1}=\frac{p-(\vec{a} \cdot \vec{n})}{\vec{b} \cdot \vec{n}}\)
Substuting (3) in (1), we get
\(\vec{u}=\vec{a}+\left(\frac{p-(\vec{a} \cdot \vec{n})}{\vec{b} \cdot \vec{n}}\right) \vec{b}, \vec{b} \cdot \vec{n} \neq 0\)

Consider the equation
\(\left(\vec{r} \cdot \vec{n}_{1}-d_{1}\right)+\lambda\left(\vec{r} \cdot \vec{n}_{2}-d_{2}\right)=0\)     .(1)
The above equation can be simplified as
\(\vec{r} \cdot\left(\vec{n}_{1}+\lambda \vec{n}_{2}\right)-\left(d_{1}+\lambda d_{2}\right)=0\)       .(2)
\(\text { Put } \vec{n}=\vec{n}_{1}+\lambda \vec{n}_{2}, d=\left(d_{1}+\lambda d_{2}\right)\)
Then the equation (2) becomes
\(\vec{r} \cdot \vec{n}=d\)         ...(3)
The equation (3) represents a plane. Hence (1) represents a plane
Let \(\overrightarrow{r_{1}}\) be the position vector of any point on the line of intersection of the plane. Then \(\overrightarrow{r_{1}}\)satisfies both the equations \(\vec{r} \cdot \vec{n}_{1}=d_{1} \text { and } \vec{r} \cdot \vec{n}_{2}=d_{2}\)  So, we have
\( \overrightarrow{r_{1}} \cdot \vec{n}_{1} =d_{1} \)    .(4)
\(\text { and } \overrightarrow{r_{2}} \cdot \vec{n}_{2} =d_{2}\)      ...(5)
By (4) and (5), \(\overrightarrow{r_{1}}\) satisfies (1). So, any point on the line of intersection lies on the plane (1). This proves that the plane (1) passes through the line of intersection.
The cartesian equation of a plane which passes through the line of intersection of the planes \(a_{1} x+b_{1} y+c_{1} z=d_{1} \text { and } a_{2} x+b_{2} y+c_{2} z=d_{2}\) is given by
\(\left(a_{1} x+b_{1} y+c_{1} z-d_{1}\right)+\lambda\left(a_{2} x+b_{2} y+c_{2} z-d_{2}\right)=0\) 

Let A (x₁ , y₁ , z₁) be any point on the plane  ax + by + cz + d₂ = 0 , then we have
\(a x_{1}+b y_{1}+c z_{1}+d_{2}=0 \Rightarrow a x_{1}+b y_{1}+c z_{1}=-d_{2}\)
The distance of the plane  ax + by + cz + d₁ = 0 from the point A (x₁ , y₁ , z₁) is given by
\(\delta=\frac{\left|a x_{1}+b y_{1}+c z_{1}+d_{1}\right|}{\sqrt{a^{2}+b^{2}+c^{2}}}=\frac{\left|d_{1}-d_{2}\right|}{\sqrt{a^{2}+b^{2}+c^{2}}}\)
Hence, the distance between two parallel planes ax + by + cz + d₁ = 0 and ax + by + cz + d₂ = 0 is given by \(\delta=\frac{\left|d_{1}-d_{2}\right|}{\sqrt{a^{2}+b^{2}+c^{2}}}\)

Let A be the point whose position vector is \(\vec{u}\)
Let F be the foot of the perpendicular from the point A to the plane \(\vec{r} \cdot \vec{n}=p\). The line joining F and A is parallel to the normal vector \(\vec{F}\) and hence its equation is \(\vec{r}=\vec{u}+t \vec{n}\).
But F is the point of intersection of the line \(\vec{r}=\vec{u}+t \vec{n}\) and the given plane \(\vec{r} \cdot \vec{n}=p\) . If \(\overrightarrow{r_{1}}\) is the position vector of F, then \(\overrightarrow{r_{1}}=\vec{u}+t_{1} \vec{n}\) for some \(t_{1} \in \mathbb{R}, \text { and } \vec{r}_{1} \cdot \vec{n}=p\) Eliminating \(\overrightarrow{r_{1}}\) we get 
\(\left(\vec{u}+t_{1} \vec{n}\right) \cdot \vec{n}=p\) which implies \(t_{1}=\frac{p-(\vec{u} \cdot \vec{n})}{|\vec{n}|^{2}} \mid\)
Now, \(\overrightarrow{F A}=\vec{u}-\left(\vec{u}+t_{1} \vec{n}\right)=-t_{1} \vec{n}=\left(\frac{(\vec{u} \cdot \vec{n})-p}{|\vec{n}|^{2}}\right) \vec{n}\)
Therefore, the length of the perpendicular from the point A to the given plane is
\(\delta=|\overrightarrow{F A}|=\left|\left(\frac{(\vec{u} \cdot \vec{n})-p}{|\vec{n}|^{2}}\right) \vec{n}\right|=\left|\frac{(\vec{u} \cdot \vec{n})-p}{|\vec{n}|}\right|\)
The position vector of the foot F of the perpendicular AF is given by
\( \overrightarrow{r_{1}}=\vec{u}+t_{1} \vec{n} \text { or } \)
\(\overrightarrow{r_{1}}=\vec{u}+\left(\frac{p-\vec{u} \cdot \vec{n}}{|\vec{n}|^{2}}\right) \vec{n}\)  

If θ is the acute angle between two planes \(\vec{r} \cdot \vec{n}_{1}=p_{1} \text { and } \vec{r} \cdot \vec{n}_{2}=p_{2}\) then \(\theta\) is the acute angle between their normal vectors \(\vec{n}_{1} \text { and } \vec{n}_{2}\) .
Therefore, \(\cos \theta=\left(\frac{\left|\vec{n}_{1} \cdot \vec{n}_{2}\right|}{\left|\vec{n}_{1}\right|\left|\vec{n}_{2}\right|}\right) \Rightarrow \theta=\cos ^{-1}\left(\frac{\left|\vec{n}_{1} \cdot \vec{n}_{2}\right|}{\left|\vec{n}_{1}\right|\left|\vec{n}_{2}\right|}\right)\)

If \(\vec{n}_{1} \text { and } \vec{n}_{2}\) are the vectors normal to the two given planes \(a_{1} x+b_{1} y+c_{1} z+d_{1}=0\) and \(a_{2} x+b_{2} y+c_{2} z+ d_{2}=0\) respectively. Then \(\vec{n}_{1}=a_{1} \hat{i}+b_{1} \hat{j}+c_{1} \hat{k} \text { and } \vec{n}_{2}=a_{2} \hat{i}+b_{2} \hat{j}+c_{2} \hat{k}\)
Therefore, using equation (1) the acute angle \(\theta\) between the planes is given by
\(\theta=\cos ^{-1}\left(\frac{\left|a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}\right|}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}} \sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}\right)\)

Consider a plane passing through three non-collinear points A, B,C with position vectors \(\vec{a}, \vec{b}, \vec{c}\) respectively. Then atleast two of them are non-zero vectors. Let us take \(\vec{b} \neq \overrightarrow{0} \text { and } \vec{c} \neq \overrightarrow{0} \text {. Let } \vec{r}\) be the position vector of an arbitrary point P on the plane. Take a point D on AB (produced) such that \(\overrightarrow{A D}\) is parallel to \(\overrightarrow{A B} \text { and } \overrightarrow{D P}\) is parallel to \(\overrightarrow{A C}\) Therefore,
\(\overrightarrow{A D}=s(\vec{b}-\vec{a}), \overrightarrow{D P}=t(\vec{c}-\vec{a})\)
Now, in triangle ADP, we have
\(\overrightarrow{A P}=\overrightarrow{A D}+\overrightarrow{D P} \text { or } \vec{r}-\vec{a}=s(\vec{b}-\vec{a})+t(\vec{c}-\vec{a}), \text { where } \vec{b} \neq \overrightarrow{0}, \vec{c} \neq \overrightarrow{0} \text { and } s, t \in \mathbb{R}\)
That is, \(\vec{r}=\vec{a}+s(\vec{b}-\vec{a})+t(\vec{c}-\vec{a})\)
This is the parametric form of vector equation of the plane passing through the given three non-collinear points

The equation ax + by + cz + d = 0 can be written in the vector form as follows
\((x \hat{i}+y \hat{j}+z \hat{k}) \cdot(a \hat{i}+b \hat{j}+c \hat{k})=-d \ \text { or } \ \vec{r} \cdot \vec{n}=-d\) 
Since this is the vector form of the equation of a plane in standard form, the given equation ax + by + cz + d = 0 represents a plane.
Here \(\vec{n}=a \hat{i}+b \hat{j}+c \hat{k}\) is a vector normal to the plane.

Consider a plane whose perpendicular distance from the origin is p .
Let A be the foot of the perpendicular from O to the plane.
Let \(\hat d\) be the unit normal vector in the direction of \(\overrightarrow{O A}\)
Then \(\overrightarrow{O A}=p \hat{d}\)
If \(\vec{r}\) is the position vector of an arbitrary point P on the plane then \(\overrightarrow{A P}\) is perpendicular to \(\overrightarrow{O A}\)
Therefore, \(\overrightarrow{A P} \cdot \overrightarrow{O A}=0 \Rightarrow(\vec{r}-p \hat{d}) \cdot p \hat{d}=0\)
\(\Rightarrow(\vec{r}-p \hat{d}) \cdot \hat{d}=0\)
which gives \(\vec{r} \cdot \hat{d}=p\)
The above equation is called the vector equation of the plane in normal form.

The two skew lines \(\vec{r}=\vec{a}+s \vec{b} \text { and } \vec{r}=\vec{c}+t \vec{d}\) are denoted by L₁ and  L₂ respectively.
Let A and C be the points on  L₁ and  L₂ with position vectors \(\vec{a} \text { and } \vec{c}\) respectively.
From the given equations of skew lines, we observe that L₁ is parallel to the vector \(\vec{b}\) and  L₂ is parallel to the vector \(\vec{d}\) So, \(\vec{b} \times \vec{d}\) is perpendicular to the lines L₁ and  L₂ .
Let SD be the line segment perpendicular to both the lines  L₁ and  L₂ . Then the vector \(\overrightarrow{S D} \) is perpendicular to the vectors \(\vec{b}\) and \(\vec{d}\) and therefore it is parallel to the vector \(\vec{b} \times \vec{d}\) .
So, \(\frac{\vec{b} \times \vec{d}}{|\vec{b} \times \vec{d}|}\) is a unit vector in the direction of \(\overrightarrow{S D} \) Then, the shortest distance \(|\overrightarrow{S D}|\) is the absolute value of the projection of \(\overrightarrow{AC} \) and \(\overrightarrow{S D} \). That is
\(\delta=|\overrightarrow{S D}|=\mid \overrightarrow{A C}\) (Unit vector in the direction of \(\overrightarrow{S D})|=|(\vec{c}-\vec{a}) \cdot \frac{\vec{b} \times \vec{d}}{|\vec{b} \times \vec{d}|} \mid\)
\(\delta=\frac{|(\vec{c}-\vec{a}) \cdot(\vec{b} \times \vec{d})|}{|\vec{b} \times \vec{d}|}, \text { where }|\vec{b} \times \vec{d}| \neq 0\)

The given two parallel lines \(\vec{r}=\vec{a}+s \vec{b} \text { and } \vec{r}=\vec{c}+t \vec{b}\) are denoted by  L₁ and  L₂ respectively. Let A and B be the points on  L₁ and  L₂ whose position vectors are \(\vec{a} \text { and } \vec{c}\) respectively. The two given lines are parallel to \(\vec{b}\) .
Let AD be a perpendicular to the two given lines. If \(\theta\) is the acute angle between \(\overrightarrow{A B} \text { and } \vec{b}\) then
\(\sin \theta=\frac{|\overrightarrow{A B} \times \vec{b}|}{|\overrightarrow{A B}||\vec{b}|}=\frac{|(\vec{c}-\vec{a}) \times \vec{b}|}{|\vec{c}-\vec{a}||\vec{b}|}\)    ....(1)
But, from the right angle triangle ABD,
\(\sin \theta=\frac{d}{A B}=\frac{d}{|\overrightarrow{A B}|}=\frac{d}{|\vec{c}-\vec{a}|}\)      ..(2)
From (1) and (2), we have \(d=\frac{|(\vec{c}-\vec{a}) \times \vec{b}|}{|\vec{b}|}, \text { where }|\vec{b}| \neq 0\)

parametric vector form of line
\(\vec{r}=\vec{a}+t \vec{n}\)
where a is position vector and n is normal vector
line from position vector a and b
normal vector \(\vec{n} = \vec{b}-\vec{a}\)
eq of line \(\vec{r}=\vec{a}+t(\vec{b}-\vec{a})\)

If \(\vec{a} \) is the position vector of a given point A and \(\vec{r} \) position vector of an arbitrary point P on the straight line, then \(\overrightarrow{A P}=\vec{r}-\vec{a}\)
Since \(\overrightarrow{A P}\) is parallel to \(\vec{b} \) we have
\( \vec{r}-\vec{a} =t \vec{b}, t \in \mathbb{R} \)  ...........(1)
\(\text {Or } \ \ \vec{r} =\vec{a}+t \vec{b}, t \in \mathbb{R}\)    ............(2) 
This is the vector equation of the straight line in parametric form

Since dot and cross can be interchanged in a scalar product, we get
\((\vec{a} \times \vec{b}) \cdot(\vec{c} \times \vec{d})=\vec{a} \cdot(\vec{b} \times(\vec{c} \times \vec{d}))\)
\(=\vec{a} \cdot((\vec{b} \cdot \vec{d}) \vec{c}-(\vec{b} \cdot \vec{c}) \vec{d})\) (by vector triple product expansion)
\(=(\vec{a} \cdot \vec{c})(\vec{b} \cdot \vec{d})-(\vec{a} \cdot \vec{d})(\vec{b} \cdot \vec{c})\)
\(=\left|\begin{array}{ll} \vec{a} \cdot \vec{c} & \vec{a} \cdot \vec{d} \\ \vec{b} \cdot \vec{c} & \vec{b} \cdot \vec{d} \end{array}\right|\)

Using vector triple product expansion, we have
\( \vec{a} \times(\vec{b} \times \vec{c})=(\vec{a} \cdot \vec{c}) \vec{b}-(\vec{a} \cdot \vec{b}) \vec{c} \ \)
\(\vec{b} \times(\vec{c} \times \vec{a})=(\vec{b} \cdot \vec{a}) \vec{c}-(\vec{b} \cdot \vec{c}) \vec{a}\)
\(\vec{c} \times(\vec{a} \times \vec{b})=(\vec{c} \cdot \vec{b}) \vec{a}-(\vec{c} \cdot \vec{a}) \vec{b}\)
Adding the above equations and using the scalar product of two vectors is commutative, we get \(\vec{a} \times(\vec{b} \times \vec{c})+\vec{b} \times(\vec{c} \times \vec{a})+\vec{c} \times(\vec{a} \times \vec{b})=\overrightarrow{0}\)

Let us choose the coordinate axes as follows :
Let x -axis be chosen along the line of action of  \(\vec{a}\) y -axis be chosen in the plane passing through \(\vec{a}\) and parallel to \(\vec{b}\) and z -axis be chosen perpendicular to the plane containing \(\vec{a}\) and \(\vec{b}\). Then, we have
\( \vec{a}=a_{1} \hat{i} \)
\(\vec{b}=b_{1} \hat{i}+b_{2} \hat{j} \)
\(\vec{c}=c_{1} \hat{i}+c_{2} \hat{j}+c_{3} \hat{k}\)
Now, \(\vec{a} \times(\vec{b} \times \vec{c})=a_{1} \hat{i} \times\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ b_{1} & b_{2} & 0 \\ c_{1} & c_{2} & c_{3} \end{array}\right|\)
\( =a_{1} \hat{i} \times\left(b_{2} c_{3} \hat{i}-b_{1} c_{3} \hat{j}+\left(b_{1} c_{2}-b_{2} c_{1}\right) \hat{k}\right) \ \)
\(=-a_{1} b_{1} c_{3} \hat{k}+a_{1}\left(b_{2} c_{1}-b_{1} c_{2}\right) \hat{j}\)   ....... (1)    \( (\vec{a} \cdot \vec{c}) \vec{b}-(\vec{a} \cdot \vec{b}) \vec{c} =a_{1} c_{1} \times\left(b_{1} \hat{i}+b_{2} \hat{j}\right)-a_{1} b_{1}\left(c_{1} \hat{i}+c_{2} \hat{j}+c_{3} \hat{k}\right) \)
\(=a_{1}\left(b_{2} c_{1}-b_{1} c_{2}\right) \hat{j} -a_{1} b_{1} c_{3} \hat{k}\)  ..... (2)
From equations (1) and (2) we get
\(\vec{a} \times(\vec{b} \times \vec{c})=(\vec{a} \cdot \vec{c}) \vec{b}-(a \cdot \vec{b}) \vec{c}\)

\(\vec a = 2\vec i + 2\vec j+ 3\vec k, \vec n = 3\vec i + 4\vec j + 3\vec k\)
vector equation
\(\bar r. \bar n = \bar a. \bar n\)
\(\bar r.(3\vec i+4\vec j+3\vec k) = (2\vec i+2\vec j+3\vec k).(3\vec i+4\vec j+3\vec k)\)
= 6+8+9
\( \vec r.\ (3\vec i+4\vec j+3\vec k)\) = 23

If \(\vec{a}, \vec{b}, \vec{c}\) are non-coplanar and
\(\left|\begin{array}{lll} x_{1} & y_{1} & z_{1} \\ x_{2} & y_{2} & z_{2} \\ x_{3} & y_{3} & z_{3} \end{array}\right| \neq 0\)
then the three vectors \(\vec{p}=x_{1} \vec{a}+y_{1} \vec{b}+z_{1} \vec{c}, \quad \vec{q}=x_{2} \vec{a}+y_{2} \vec{b}+z_{2} \vec{c}, \text { and }, \vec{r}=x_{3} \vec{a}+y_{3} \vec{b}+z_{3} \vec{c}\) are also non-coplanar.

Let \(\vec{a}=a_{1} \hat{i}+a_{2} \hat{j}+a_{3} \hat{k}, \vec{b}=b_{1} \hat{i}+b_{2} \hat{j}+b_{3} \hat{k}, \vec{c}=c_{1} \hat{i}+c_{2} \hat{j}+c_{3} \hat{k}\) Then, we have
\(\vec{a}, \vec{b}, \vec{c}\) are coplanar \(\Leftrightarrow[\vec{a}, \vec{b}, \vec{c}]=0 \Leftrightarrow\left|\begin{array}{lll} a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \\ c_{1} & c_{2} & c_{3} \end{array}\right|=0\)
there exist scalars r, s, t \(\in \mathbb{R}\)
atleast one of them non-zero such that
\(a_{1} r+a_{2} s+a_{3} t=0, \quad b_{1} r+b_{2} s+b_{3} t=0, c_{1} r+c_{2} s+c_{3} t=0\)
there exist scalars r, s, t \(\in \mathbb{R}\)
atleast one of them non-zero such that \(r \vec{a}+s \vec{b}+t \vec{c}=\overrightarrow{0}\)

Let \(\vec{a}, \vec{b}, \vec{c}\) be any three non-zero vectors. Then,
 \((\vec{a} \times \vec{b}) \cdot \vec{c}=0 \Leftrightarrow \vec{c}\) is perpendicular to \(\vec{a} \times \vec{b}\)
\(\Leftrightarrow \vec{c}\) lies in the plane which is parallel to both \(\vec{a} \text { and } \vec{b}\)
\(\Leftrightarrow \vec{a}, \vec{b}, \vec{c}\) are coplanar.

Let \(\vec{a}, \vec{b}, \vec{c}\) be any three non-zero vectors. Then,
\((\vec{a} \times \vec{b}) \cdot \vec{c}=0 \Leftrightarrow \vec{c}\) is perpendicular to \(\vec{a} \times \vec{b}\)
\(\Leftrightarrow \vec{c}\) lies in the plane which is parallel to both \(\vec{a} \text { and } \vec{b}\)
\(\Leftrightarrow \vec{a}, \vec{b}, \vec{c}\) are coplanar.

Using the properties of scalar product and vector product, we get
\( {[(\vec{a}+\vec{b}), \vec{c}, \vec{d}] } =((\vec{a}+\vec{b}) \times \vec{c}) \cdot \vec{d} \)
\(=(\vec{a} \times \vec{c}+\vec{b} \times \vec{c}) \cdot \vec{d} \)
\(=(\vec{a} \times \vec{c}) \cdot \vec{d}+(\vec{b} \times \vec{c}) \cdot \vec{d} \)
\(=[\vec{a}, \vec{c}, \vec{d}]+[\vec{b}, \vec{c}, \vec{d}] \)
\({[\lambda \vec{a}, \vec{b}, \vec{c}] } =((\lambda \vec{a}) \times \vec{b}) \cdot \vec{c}=(\lambda(\vec{a} \times \vec{b})) \cdot \vec{c}=\lambda((\vec{a} \times \vec{b}) \cdot \vec{c})=\lambda[\vec{a}, \vec{b}, \vec{c}] \)
Using the first statement of this result, we get the following.
\( {[\vec{a},(\vec{b}+\vec{c}), \vec{d}] } =[(\vec{b}+\vec{c}), \vec{d}, \vec{a}]=[\vec{b}, \vec{d}, \vec{a}]+[\vec{c}, \vec{d}, \vec{a}] \)
\(=[\vec{a}, \vec{b}, \vec{d}]+[\vec{a}, \vec{c}, \vec{d}] \)
\({[\vec{a}, \lambda \vec{b}, \vec{c}] } =[\lambda \vec{b}, \vec{c}, \vec{a}]=\lambda[\vec{b}, \vec{c}, \vec{a}]=\lambda[\vec{a}, \vec{b}, \vec{c}] . \)
Similarly, the remaining equalities are proved.

Let \(\vec{a}=a_{1} \hat{i}+a_{2} \hat{j}+a_{3} \hat{k}, \quad \vec{b}=b_{1} \hat{i}+b_{2} \hat{j}+b_{3} \hat{k} \text { and } \vec{c}=c_{1} \hat{i}+c_{2} \hat{j}+c_{3} \hat{k}\)
Then, \(\vec{a} \cdot(\vec{b} \times \vec{c})=(\vec{b} \times \vec{c}) \cdot \vec{a}=\left|\begin{array}{ccc} b_{1} & b_{2} & b_{3} \\ c_{1} & c_{2} & c_{3} \\ a_{1} & a_{2} & a_{3} \end{array}\right|=-\left|\begin{array}{ccc} a_{1} & a_{2} & a_{3} \\ c_{1} & c_{2} & c_{3} \\ b_{1} & b_{2} & b_{3} \end{array}\right|, \text { by } R_{1} \leftrightarrow R_{3}\)
\(=\left|\begin{array}{lll} a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \\ c_{1} & c_{2} & c_{3} \end{array}\right|, \text { by } R_{2} \leftrightarrow R_{3}\)
\(=(\vec{a} \times \vec{b}) \cdot \vec{c}\)
Hence the theorem is proved.

By definition, we have
\((\vec{a} \times \vec{b}) \cdot \vec{c}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \end{array}\right| \cdot \vec{c}\)
\( =\left[\left(a_{2} b_{3}-a_{3} b_{2}\right) \hat{i}-\left(a_{1} b_{3}-a_{3} b_{1}\right) \hat{j}+\left(a_{1} b_{2}-a_{2} b_{1}\right) \hat{k}\right] \cdot\left(c_{1} \hat{i}+c_{2} \hat{j}+c_{3} \hat{k}\right) \)
\(=\left(a_{2} b_{3}-a_{3} b_{2}\right) c_{1}+\left(a_{3} b_{1}-a_{1} b_{3}\right) c_{2}+\left(a_{1} b_{2}-a_{2} b_{1}\right) c_{3}\)
\(=\left|\begin{array}{lll} a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \\ c_{1} & c_{2} & c_{3} \end{array}\right|\)
which completes the proof of the theorem.

Here, \(\vec { a } =(2\hat { i } -\hat { j } +2\hat { k } ),\vec { b } =(3\hat { i } +4\hat { j } +2\hat { k } )\).
The vector form of the given plane is \(\vec { r } .(\hat { i } -\hat { j } +\hat { k } )=5\). Then \(\vec { r } .(\hat { i } -\hat { j } +\hat { k } )=5\) and p = 5
We know that the position vector of the point of intersection of the line  \(\vec { r } =\vec { a } +t\vec { b } \) and the plane
\(\vec { r } .\vec { d } =p\vec { u } =\vec { a } +\left( \frac { p-(\vec { a } .\vec { n } ) }{ \vec { b } .\vec { n } } \right) \vec { b } \), where \(\vec { b } .\vec { n } \neq \vec { 0 } \)
Clearly, we observe that \(\vec { b } .\vec { n } \neq \vec { 0 } \)
Now, \(\frac { p-(\vec { a } .\vec { n } ) }{ \vec { b } .\vec { n } } =\frac { 5-(2\hat { i } -\hat { j } +2\hat { k } ).(\hat { i } -\hat { j } +\hat { k } ) }{ (3\hat { i } +4\hat { j } +2\hat { k } ).(\hat { i } -\hat { j } +\hat { k } ) } =0\). Therefore, the position vector of the point of intersection of the given line and the given plane is
\(\hat { r } =(2\hat { i } -\hat { j } +2\hat { k } )+(0)(3\hat { i } +4\hat { j } +2\hat { k } )=2\hat { i } -\hat { j } +2\hat { k } \)
That is, the given straight line intersects the plane at the point (2, −1, 2)
Aliter:
The Cartesian equation of the given straight line is \(\frac { x-2 }{ 3 } =\frac { y+1 }{ 4 } =\frac { z-2 }{ 2 } =t\)(say)
We know that any point on the given straight line is of the form (3t+2, 4 t−1, 2 t+2). If the given line and the plane intersects, then this point lies on the given pane x−y+z−5 = 0.
So, (3t + 2)−(4t − 1) + (2t + 2) − 5 = 0 ⇒ t = 0.
Therefore, the given line intersects the given plane at the point (2, -1, 2)

Let \(\vec { \mu } \)  the position vector of an arbitrary point on the plane \(\vec { r } .(2\hat { i } -\hat { j } -2\hat { k } )\) = 6. Then, we have
\(\vec { \mu } .(2\hat { i } -\hat { j } -2\hat { k } )=6\) .........(1)
If δ is the distance between the given planes, then δ is the perpendicular distance from  \(\vec { \mu } \) to the plane
\(\vec { r } .(6\hat { i } -3\hat { j } -6\hat { k} )\) = 27
Therefore, δ = \(\frac { \left| \vec { u } .\vec { n } -p \right| }{ \left| \vec { n } \right| } =\left| \frac { \vec { u } .(6\hat { i } -3\hat { j } -6\hat { k } )-27 }{ \sqrt { 6^{ 2 }+(-{ 3) }^{ 2 }+(-{ 6 })^{ 2 } } } \right| =\left| \frac { 3(\vec { u } .(2\hat { i } -\hat { j } -2\hat { k } ))-27 }{ 9 } \right| =\left| \frac { (3(6)-27 }{ 9 } \right| =1\) unit

\(\hat { d } =\frac { 3\hat { i } -4\hat { j } +5\hat { k } }{ \sqrt { { 3 }^{ 2 }+(-4)+{ 5 }^{ 2 } } } =\frac { 3\hat { i } -4\hat { j } +5\hat { k } }{ \sqrt { 9+16+25 } } \)
\(=\frac { 3\hat { i } -4\hat { j } +5\hat { k } }{ \sqrt { 50 } } =\frac { 3\hat { i } -4\hat { j } +5\hat { k } }{ 5\sqrt { 2 } } \)
[∵ 3, -4, 5 are direction ratios]. The equation of the plane at a distance p from the origin and perpendicular to the unit normal vector \(\hat { d }\ is\ \vec { r } .\hat { d } =p\)
Equation of the required plane is
\(\vec { r } .\left( \frac { 3\hat { i } -4\hat { j } +5\hat { k } }{ 5\sqrt { 2 } } \right) =7\)

Let \(\vec { d } =3\hat { i } -4\hat { j } +12\hat { k } \) and q = 5.
If \(\vec { d } \) is the unit vector in the direction of the vector \(3\hat { i } -4\hat { j } +12\hat { k } \), then \(\vec { d } =\frac { 1 }{ 13 } (3\hat { i } -4\hat { j } +12\hat { k } )\)
Now, dividing the given equation by 13, we get
\(\hat { r } .\left( \frac { 3 }{ 13 } \hat { i } -\frac { 4 }{ 13 } \hat { j } +\frac { 12 }{ 13 } \hat { k } \right) =\frac { 5 }{ 13 } \)
which is the equation of the plane in the normal form \(\hat { r } .\hat { d } =p\)
From this equation, we infer that \(\hat { d } =\frac { 1 }{ 3 } (3\hat { i } -4\hat { j } +12\hat { k } )\) is a unit vector normal to the plane from the origin. Therefore, the direction cosines of \(\frac { 3 }{ 13 } ,\frac { -4 }{ 13 } ,\frac { 12 }{ 13 } \) and the length of the perpendicular from the origin to the plane is \(\frac { 5 }{ 13 } \)

The parametric form of vector equations of the given straight lines are
\(\vec { r } =(2\hat { i } +3\hat { j } +4\hat { k } )+t(-2\hat { i } +\hat { j } -2\hat { k } )\)
and \(\vec { r } =(3\hat { i } -2\hat { k } )+t(2\hat { i } -\hat { j } +2\hat { k } )\)
Comparing the given two equations with \(\vec { r } =\vec { a } +t\vec { b } ,\vec { r } =\vec { c } +s\vec { d } \)
we have \(\vec { a } =2\hat { i } +3\hat { j } +4\hat { k } ,\vec { b } =-2\hat { i } +\hat { j } -2\hat { k } ,\vec { c } =3\hat { i } -2\hat { k } ,\vec { d } =2\hat { i } -\hat { j } +2\hat { k } \)
Clearly, \(\vec { b } \) is a scalar multiple of \(\vec { d } \), and hence the two straight lines are parallel. We know that the shortest distance between two parallel straight lines is given by \(d=\frac { \left| (\vec { c } -\vec { a } )\times \vec { b } \right| }{ \left| \vec { b } \right| } \)
\((\vec { c } -\vec { a } )\times \vec { b } =\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 1 & -3 & -6 \\ -2 & 1 & -2 \end{matrix} \right| =12\vec { i } +14\vec { j } -5\vec { k } \)
\(d=\frac { \left| 12\hat { i } +14\hat { j } -5\hat { k } \right| }{ \left| -2\hat { i } +\hat { j } -2\hat { k } \right| } =\frac { \sqrt { 365 } }{ 3 } \)

The straight line passing through the points (6, 7, 5)A and (8, 10, 6)B is parallel to the vector \(\vec { b } =\vec { AB } =\vec { OB } -\vec { OA } =2\hat { i } +3\hat { j } +\hat { k } \) and the straight line passing through the points C(10, 2, -5) and D(8, 3, -4) is parallel to the vector \(\vec { d } =\vec { CD } =-2\hat { i } +\hat { j } +\hat { k } \). Therefore, the angle between the two straight lines is the angle between the two vectors \(\vec { b } \)and \(\vec { d } \).
Since \(\vec { b } .\vec { d } =(2\hat { i } +3\hat { j } +\hat { k).( } -2\hat { i } +\hat { j } +\hat { k } ) =0\)
the two vectors are perpendicular, and hence the two straight lines are perpendicular. 
Aliter :
We find that direction ratios of the straight line joining the points A(6, 7, 5) and B(8,10, 6) are (b₁, b₂, b₃ ) = (2, 3, 1) and direction ratios of the line joining the points C(10, 2, −5) and D(8, 3, −4) are (d₁, d₂, d₃ ) = (−2, 1, 1). Since  b₁d₁+ b₂d₂ + b₃d₃ = (2)(−2) + (3)(1) + (1)(1) = 0, the two straight lines are perpendicular.

Comparing the given lines with the general Cartesian equations of straight lines,
\(\frac { x-{ x }_{ 1 } }{ { b }_{ 1 } } =\frac { y-{ y }_{ 1 } }{ { b }_{ 2 } } =\frac { z-{ z }_{ 1 } }{ { b }_{ 3 } } \)and \(\frac { x-{ x }_{ 2 } }{ { d }_{ 1 } } =\frac { y-{ y }_{ 2 } }{ { d }_{ 2 } } =\frac { z-{ z }_{ 2 } }{ { d }_{ 3 } } \)
we find (b₁, b₂, b₃) = (2, 1, -2) and (d₁, d₂, d₃) = (4, -4, 2). Therefore, the angle between the two straight lines is 
\(\theta ={ cos }^{ -1 }\left( \frac { \left| (2)(4)+(1)(-4)+(-2)(2) \right| }{ \sqrt { { 2 }^{ 2 }+{ 1 }^{ 2 }+({ -2) }^{ 2 } } \sqrt { { 4 }^{ 2 }+(-{ 4 })^{ 2 }+{ 2 }^{ 2 } } } \right) ={ cos }^{ -1 }\left( 0 \right) =\frac { \pi }{ 2 } \)
Thus the two straight lines are perpendicular.

We know that the line \(\vec { r } =(\hat { i } +2\hat { j } +4\hat { k } )+t(2\hat { i } +2\hat { j } +\hat { k } )\) is parallel to the vector \(2\hat { i } +2\hat { j } +\hat { k } \).
Direction ratios of the straight line joining the two given points (5, 1, 4) and (9, 2, 12) are 4,1,8 and hence this line is parallel to the vector \(\hat { 4i } +\hat { j } +8\hat { k } \)
Therefore, the angle between the given two straight lines is
\(\theta ={ cos }^{ -1 }\left( \frac { \left| \vec { b } .\vec { d } \right| }{ \left| \vec { b } \right| \left| \vec { d } \right| } \right) \), where \(​​\vec { b } \) = \(2\hat { i } +2\hat { j } +\hat { k } \) and \(\vec { d } \) = \(\hat { 4i } +\hat { j } +8\hat { k } \)
Therefore, \(\theta ={ cos }^{ -1 }\left( \frac { \left| (2\hat { i } +2\hat { j } +\hat { k } ).(4\hat { i } +\hat { j } +8\hat { k } ) \right| }{ \left| 2\hat { i } +2\hat { j } +\hat { k } \right| \left| 4\hat { i } +\hat { j } +8\hat { k } \right| } \right) ={ cos }^{ -1 }\left( \frac { 2 }{ 3 } \right) \)

If \(\hat { b } =\frac { 2\hat { i } +2\hat { j } -\hat { k } }{ \left| \hat { 2i } +2\hat { j } -\hat { k } \right| } =\frac { 1 }{ 3 } (2\hat { i } +2\hat { j } -\hat { k } )\). Therefore from the definition of direction cosines of \(\hat { b } \), we have
\(cos\alpha =\frac { 2 }{ 3 } ,cos\beta =\frac { 2 }{ 3 } ,cos\gamma =-\frac { 1 }{ 3 } \)
where α, β, \(\gamma \) are the angles made by \(\hat { b } \) with the positive x -axis, positive y -axis, and positive
z -axis, respectively. As the angle between the given straight line with the coordinate axes are same as the angles made by \(\hat { b } \) with the coordinate axes, we have \(\alpha ={ cos }^{ -1 }\left( \frac { 2 }{ 3 } \right) ,\beta ={ cos }^{ -1 }\left( \frac { 2 }{ 3 } \right) ,\gamma ={ cos }^{ -1 }\left( \frac { -1 }{ 3 } \right) \) respectively.

Treating \((\vec { a } \times \vec { b } )\) as the first vector on the right hand side of the given equation and using the vector triple product expansion, we get
\((\vec { a } \times \vec { b } )\times (\vec { a } \times \vec { c } )=((\vec { a } \times \vec { b } ).\vec { c } )\vec { a } -((\vec { a } \times \vec { b } ).\vec { a } )\vec { c } =(\vec { a } .\vec { b } \times \vec { c } ))\vec { a } \)

Using the definition of the scalar triple product, we get 
\([\vec { a } \times \vec { b } ,\vec { b } \times \vec { c } ,\vec { c } \times \vec { a } ]\) = \((\vec { a } \times \vec { b } ).[(\vec { b } \times \vec { c } )\times (\vec { c } \times \vec { a } )]\) ....(1)
By treating \((\vec { b } \times \vec { c } )\) as the first vector in the vector triple product, we find
\((\vec { b } \times \vec { c } )\times (\vec { c } \times \vec { a } )\) = \(((\vec { b } \times \vec { c } ).\vec { a } )\vec { c } \) -  \(((\vec { b } \times \vec { c } ).\vec { c } )\vec { a } )\) = \([{ \vec { a } ,\vec { b } ,\vec { c } }]\vec { c } \)
Using this value in (1), we get
\([\vec { a } \times \vec { b } ,\vec { b } \times \vec { c } ,\vec { c } \times \vec { a } ]\) = \((\vec { a } \times \vec { b } ).([\vec { a } ,\vec { b } ,\vec { c } ]\vec { c } )=[\vec { a } ,\vec { b } ,\vec { c } ](\vec { a } \times \vec { b } ).\vec { c } ={ [\vec { a } ,\vec { b } ,\vec { c } ] }^{ 2 }\)

We get \([\vec { a } +\vec { c } ,\vec { a } +\vec { b } ,\vec { a } +\vec { b } +\vec { c } ]\) = \(\left| \begin{matrix} 1 & 0 & 1 \\ 1 & 1 & 0 \\ 1 & 1 & 1 \end{matrix} \right| \)\([\vec { a } ,\vec { b } ,\vec { c } ]\) = \([\vec { a } ,\vec { b } ,\vec { c } ]\)

Given \(\vec { F } =3\hat { i } +4\hat { j } -5\hat { k } \)
\(\vec { r } \) = (Force acting through the point) - (force acting to the point)
= \((4\hat { i } +2\hat { j } -3\hat { k } )-(2\hat { i } -\hat { j } +4\hat { k } )\)
= \(2\hat { i } +5\hat { j } -7\hat { k } \)
Torque = \(\vec { c } =\hat { r } \times \hat { F } \)
=  \(\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 2 & 5 & -7 \\ 3 & 4 & -5 \end{matrix} \right| =\hat { i } \left| \begin{matrix} 5 & -7 \\ 4 & -5 \end{matrix} \right| -\hat { j } \left| \begin{matrix} 2 & -7 \\ 3 & -5 \end{matrix} \right| +\hat { k } \left| \begin{matrix} 2 & 5 \\ 3 & 4 \end{matrix} \right| \)
\(\hat { i } =\hat { i } (-25+28)-\hat { j } (-10+21)+\hat { j } (8-15)\)
= \(3\hat { i } -11\hat { j } -7\hat { k } \)
∴ MagnitudeoftheTorque = \(\sqrt { { 3 }^{ 2 }+(-11)^{ 2 }+(-7)^{ 2 } } \)
= \(\sqrt { 9+121+49 } =\sqrt { 179 } \)
Hence, the direction cosines are \(\left( \frac { 3 }{ \sqrt { 179 } } ,\frac { -11 }{ \sqrt { 179 } } ,\frac { -7 }{ \sqrt { 179 } } \right) \).

  
Vector area of quadrilateral ABCD
= vector area of ΔABC + vector area of ΔACD
\(\frac { 1 }{ 2 } (\vec { AB } \times \vec { AC } )+\frac { 1 }{ 2 } (\vec { AC } \times \vec { AD } )\)
= \(-\frac { 1 }{ 2 } (\vec { AC } \times \vec { AB } )+\frac { 1 }{ 2 } (\vec { AC } \times \vec { AD } )\)
\(\left[ \because \ \vec { b } \times \vec { a } =-(\vec { a } \times \vec { b } ) \right] \)
= \(\frac { 1 }{ 2 } \vec { AC } \times (-\vec { AB } +\vec { AD } )\)
= \(\frac { 1 }{ 2 } \vec { AC } \times (\vec { BA } +\vec { AD } )\)  \([\because \vec { AB } =-\vec { BA } ]\)
= \(\frac { 1 }{ 2 } \vec { AC } \times \vec { BD } \) [By Δ law of addition]
∴ Area of the quadrilateral ABCD = \(\frac { 1 }{ 2 } \vec { AC } \times \vec { BD } \)

Let A be the point (2, 0, -1). Then the position vector of A is  \(\vec { OA } \) = \(2\hat { i } -\hat { k } \) and therefore \(\vec { r } =\vec { OA } =-2\hat { i } -\hat { k } \)

Then the given force is \(\vec{F}=\hat { 2i } +\hat { j } -\hat { k } \) So, the torque is 
\(\vec { t } =\vec { r } \times \vec { F } =\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ -2 & 0 & 1 \\ 2 & 1 & -1 \end{matrix} \right| =-\hat { i } -2\hat { k } \)
The magnitude of the torque = \(\left| -\hat { i } -\hat { 2k } \right| =\sqrt { 5 } \) and the direction cosines of the torque are \(-\frac { 1 }{ \sqrt { 5 } } \), 0 , \(-\frac { 2 }{ \sqrt { 5 } } \)

Resultant of the given forces is \(\hat{F}\) = ( \(\hat { 2j } +\hat { 5j } +\hat { 6k } \) )+ (\(-\hat { i } -\hat { 2j } -\hat { k } \) ) = \(\hat { i } +\hat { 3j } +\hat {5 k } \)
Let A and B be the points (4, −3, −2) and (6, 1,  −3) respectively.
Then the displacement vector of the particle is 
\(\vec { d } =\vec { AB } =\vec { OB } -\vec { OA } =(\hat { 6i } +\hat { j } -\hat { 3k } )-(\hat { 4i } -\hat { 3j } -\hat { 2k } )=\hat { 2i } +\hat { 4j } -\hat { k } \)
Therefore the work done
w = \(\vec { f } .\vec { d } =(\hat { i } +\hat { 3j } +\hat { 5k } ).(\hat { 2j } +\hat { 4j } -\hat { k } )\) = 9 units.

We know that the formula for the distance between two parallel ax + by + cz + d₁ = 0 and ax + by + cz + d₂ = 0 is \(\delta =\frac { |{ d }_{ 1 }-{ d }_{ 2 }| }{ \sqrt { { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 } } } \). Rewrite the second equation as x + 2y - 2z + \(\frac { 5 }{ 2 } \) = 0. 
Comparing the given equations with the general equations, we get a = 1, b = 2, c = -2, d₁ = 1, d₂ = \(\frac { 5 }{ 2 } \). Substituting these values in the formula, we get the distance
\(\delta =\frac { |{ d }_{ 1 }-{ d }_{ 2 }| }{ \sqrt { { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 } } } =\frac { |1-\frac { 5 }{ 2 } | }{ \sqrt { { 1 }^{ 2 }+{ 2 }^{ 2 }+(-2^{ 2 }) } } =\frac { 1 }{ 2 } \) units.

Comparing the given equation of the plane with \(\vec { r } .\vec { n } \) = p, we have \(\vec { n } =6\hat { i } -3\hat { j } +2\hat { k } \).
We know that the perpendicular distance from the given point with position vector u to the plane \(\vec { r } .\vec { n } \)= p is given by \(\delta =\frac { |\vec { u } .\vec { n } -p| }{ |\vec { n } | } \). Therefore, substi \(\vec { u } \)= (2, 5, -3) = \(2\hat { i } +5\hat { j } -3\hat { k } \) and \(\ \vec { n } =6\hat { i } -3\hat { j } +2\hat { k } \) in the formula, we get
\(\delta =\frac { |\vec { u } .\vec { n } -p| }{ |\vec { n } | } =\frac { |(2\hat { i } +5\hat { j } -3\hat { k } ).(6\hat { i } -3\hat { j } +2\hat { k } )-5| }{ |6\hat { i } -3\hat { j } +2\hat { k } | } \) = 2 unit.

The normal vectors of the two given planes \(\vec { r } .(2\hat { i } +2\hat { j } +2\hat { k } )\)= 11 and 4x+2y+2z = 15 are \(\vec { { n }_{ 1 } } =2\hat { i } +2\hat { j } +2\hat { k } \) and \(\vec { { n }_{ 2 } } =4\hat { i } -2\hat { j } +2\hat { k } \) respectively.
If θ is the acute angle between the planes, then we have
\(\theta =cos^{ -1 }\left( \frac { |\vec { { n }_{ 1 } } .\vec { { n }_{ 2 } } | }{ |\vec { { n }_{ 1 } } .\vec { { n }_{ 2 } } | } \right) =cos^{ -1 }\left( \frac { |((2\hat { i } +2\hat { j } +2\hat { k } ).(4\hat { i } -2\hat { j } +2\hat { k } ))| }{ |(2\hat { i } +2\hat { j } +2\hat { k } )||4\hat { i } -2\hat { j } +2\hat { k } | } \right) =cos^{ -1 }\left( \frac { \sqrt { 2 } }{ 3 } \right) \).

The angle between a line \(\vec { r } =\vec { a } +t\vec { b } \) and a plane \(\vec { r } .\vec { n } \) = p with normal \(\vec { n } \) is θ = \(sin^{ -1 }\left( \frac { |\vec { b } .\vec { n } | }{ |\vec { b }| .|\vec { n } | } \right) \quad \)
Here, \(\vec { b } =\hat { i } -\hat { j } +\hat { k } \) and \(\vec { n } =2\hat { i } \hat { j } +\hat { k } \)
So, we get θ = \(sin^{ -1 }\left( \frac { |\vec { b } .\vec { n } | }{ |\vec { b }| .|\vec { n } | } \right) \quad =sin^{ -1 }\left( \frac { |(\hat { i } -\hat { j } +\hat { k } ).(2\hat { i } -\hat { j } +\hat { k } )| }{ |\hat { i } -\hat { j } +\hat { k } ||2\hat { i } -\hat { j } +\hat { k } | } \right) =sin^{ -1 }\left( \frac { 2\sqrt { 3 } }{ 3 } \right) \).

Vector form of the equation of the plane is
\(\vec { r } .(6\hat { i } +4\hat { j } -3\hat { k } )\) = 12
Let \(\vec { r } =x\hat { i } +y\hat { j } +z\hat { k } \)
\(\Rightarrow (x\hat { i } +y\hat { j } +z\hat { k } ).(6\hat { i } +4\hat { j } -3\hat { k } )=12\)
⇒ 6x + 4y - 3z = 0
Dividing by 12, we get
\(\frac { 6x }{ 12 } +\frac { 4y }{ 12 } +\frac { 3z }{ 12 } =1\)
[\(\because \frac { x }{ a } +\frac { y }{ b } +\frac { z }{ c } =1\) is the equation of the plane in intercept form]
⇒ \(\frac { x }{ 2 } +\frac { y }{ 3 } +\frac { z }{ -4 } =1\)
∴ The x-intercepts of the plane is 2, y intercept is 3 and z-intercept is -4.

We observe that the straight line \(\frac { x-1 }{ 4 } =\frac { 2-y }{ 6 } =\frac { z-4 }{ 12 } \) is parallel to the vector \(4\hat { i } -6\hat { j } +12\hat { k } \) and the straight line \(\frac { x-3 }{ -2 } =\frac { y-3 }{ 3 } =\frac { 5-z }{ 6 } \) is parallel to the vector \(2\hat { i } +3\hat { j } -6\hat { k } \)
Since \(4\hat { i } -6\hat { j } +12\hat { k } =-2(-2\hat { i } +3\hat { j } -6\hat { k } )\), two vectors are parallel, and hence the two straight lines are parallel.

Given \(\vec { a } =\hat { i } -2\hat { j } +3\hat { k } ,\vec { b } =2\hat { i } +\hat { j } +\hat { k } \) and \(\vec { c } =3\hat { i } +2\hat { j } +\hat { k } \)
(i) \((\vec { a } \times \vec { b } )\) = \(\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 1 & -2 & 3 \\ 2 & 1 & -2 \end{matrix} \right| \)
= \(\hat { i } \left| \begin{matrix} -2 & 3 \\ 1 & -2 \end{matrix} \right| -j\left| \begin{matrix} 1 & 3 \\ 2 & -2 \end{matrix} \right| +\hat { k } \left| \begin{matrix} 1 & -2 \\ 2 & 1 \end{matrix} \right| \)
= \(\\ \hat { i } \)(4-3)-\(\\ \hat { j } \)(-2-6)+\(\\ \hat { k } \)(1+4)
= \(\hat { i } +8\hat { j } +5\hat { k } \)
\((\vec { a } \times \vec { b } )\times \vec { c } =\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 1 & 8 & 5 \\ 3 & 2 & 1 \end{matrix} \right| \)
=\(\hat { i } \left| \begin{matrix} 8 & 5 \\ 2 & 1 \end{matrix} \right| \hat { -j } \left| \begin{matrix} 1 & 5 \\ 3 & 1 \end{matrix} \right| +\hat { k } \left| \begin{matrix} 1 & 8 \\ 3 & 2 \end{matrix} \right| \)
 = \(\\ \hat { i } \)(8-10)-\(\\ \hat { j } \)(1-15)+\(\\ \hat { k } \)(2-24)
= \(-2\hat { i } +14\hat { j } -22\hat { k } \)
(ii) \(\vec { a } \times (\vec { b } \times \vec { c } )\)
\(\vec { b } \times \vec { c } \) = \(\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 2 & 1 & -2 \\ 3 & 2 & 1 \end{matrix} \right| \)
= \(\hat { i } \left| \begin{matrix} 1 & -2 \\ 2 & 1 \end{matrix} \right| \hat { -j } \left| \begin{matrix} 2 & -2 \\ 3 & 1 \end{matrix} \right| +\hat { k } \left| \begin{matrix} 2 & 1 \\ 3 & 2 \end{matrix} \right| \)
= \(\\ \hat { i } \)(1+4)-\(\\ \hat { j } \)(2+6)+\(\\ \hat { k } \)(4-3) = \(5\hat { i } -8\hat { j } +\hat { k } \)
∴ \(\vec { a } \times (\vec { b } \times \vec { c } )=\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 1 & -2 & 3 \\ 5 & -8 & 1 \end{matrix} \right| \)
= \(\hat { i } \left| \begin{matrix} -2 & -3 \\ -8 & 1 \end{matrix} \right| -\hat { j } \left| \begin{matrix} 1 & 3 \\ 5 & 1 \end{matrix} \right| +\hat { k } \left| \begin{matrix} 1 & -2 \\ 5 & -8 \end{matrix} \right| \)
= \(\\ \hat { i } \)(-2-24)-\(\\ \hat { j } \)(1-15)+\(\\ \hat { k } \)(-8-10)
= \(22\hat { i } +14\hat { j } +2\hat { k } \).

Let \(\vec { a } =7\hat { i } +\lambda \hat { j } -3\hat { k } ,\vec { b } =\hat { i } +2\hat { j } -\hat { k } \) and \(\vec { c } =-3\hat { i } +7\hat { j } -5\hat { k } \)
∴ volume of the parallelepiped
= \(\vec { a } .(\vec { b } \times \vec { c } )\)
Given \(\vec { a } .(\vec { b } \times \vec { c } )\) = 90
⇒ \(\left| \begin{matrix} 7 & \lambda & -3 \\ 1 & 2 & -1 \\ -3 & 7 & 5 \end{matrix} \right| \) = 90
⇒ \(-6\left| \begin{matrix} 2 & -1 \\ 7 & 5 \end{matrix} \right| -\lambda \left| \begin{matrix} 1 & -1 \\ -3 & 5 \end{matrix} \right| -3\left| \begin{matrix} 1 & 2 \\ -3 & 7 \end{matrix} \right| \) = 90
⇒ 7(10+7)-λ(5-3)-3(7+6) = 90
⇒ 7(17)-λ(2)-3(13) = 90
⇒ 119-2λ-39 = 90
⇒ 119-39-90 = 2λ
⇒ -10 = 2λ
⇒ λ = -5

Since the given three vectors are coplanar, we have \(\left| \begin{matrix} 2 & -1 & 3 \\ 3 & 2 & 1 \\ 1 & m & 4 \end{matrix} \right| \) = 0 ⇒ m = -3

Here, \(\vec { a } =\hat { i } +\hat { 2j } -\hat { 3k } \), \(\vec { b } =\hat { 2i } -\hat { j } +\hat { 2k } \), \(\vec { c } =\hat { 3i } +\hat { j } -\hat { k } \)
We know that  \(\vec { a } ,\vec { b } ,\vec { c } \)  are coplanar if and only if \([\vec { a } ,\vec { b } ,\vec { c } ]\) = 0. Now, \([\vec { a } ,\vec { b } ,\vec { c } ]\) = \(\left| \begin{matrix} 1 & 2 & -3 \\ 2 & -1 & 2 \\ 3 & 1 & -1 \end{matrix} \right| =0\)
Therefore, the three given vectors are coplanar.

We know that the volume of the parallelepiped whose coterminus edges are \(\vec { a } ,\vec { b } ,\vec { c } \) is given by |\([\vec { a } ,\vec { b } ,\vec { c } ]\)|. Here, \(\vec { a } =\hat { 2i } -\hat { 3j } +\hat { 4k } ,\vec { b } =\hat { i } +\hat { 2j } -\hat { k } ,\vec { c } =\hat { 3i } -\hat { j } +\hat { 2k } \)
Since \([\vec { a } ,\vec { b } ,\vec { c } ]\)  = \(\left| \begin{matrix} 2 & -3 & 4 \\ 1 & 2 & -1 \\ 3 & -1 & 2 \end{matrix} \right| =-7\) , the volume of the given parallelepiped is \(\left| -7 \right| =7\) cubic units.

Here, \(\vec { u } =\hat { i } +2\hat { j } +3\hat { k } ,\vec { n } =\hat { i } +2\hat { j } +4\hat { k } \), p = 38. Then the position vector of the image \(\vec { v } \)of
\(\vec { u } =\hat { i } +2\hat { j } +3\hat { k } \) is given by \(\vec { v } =\vec { u } +\frac { 2[p-(\vec { u } .\vec { n } )] }{ { \left| \vec { n } \right| }^{ 2 } } \vec { n } \)
\(\vec { v } =(\hat { i } +2\hat { j } +3\hat { k } )+\frac { 2[38-(\hat { i } +2\hat { j } +3\hat { k } ).(\hat { i } +2\hat { j } +4\hat { k } ))] }{ (\hat { i } +2\hat { j } +4\hat { k } ).(\hat { i } +2\hat { j } +4\hat { k } ) } (\hat { i } +2\hat { j } +4\hat { k } )\)
That is \(\vec { v } =(\hat { i } +2\hat { j } +3\hat { k } )+2(\frac { [38- 17]}{21}) (\hat { i } +2\hat { j } +4\hat { k } )= (3\hat { i } +6\hat { j } +11\hat { k } )\)
Therefore, the image of the point with position vector \(\hat { i } +2\hat { j } +3\hat { k } \) is \(3\hat { i } +6\hat { j } +11\hat { k } \)

The equation of the plane passing through the intersection of the planes 2x + 3y−z + 7 = 0 and x + y− 2z + 5 = 0 is (2x + 3y −z + 7) +λ (x + y −2z + 5) = 0 or (2 + λ )x+ (3 + λ )y+ (−1 − 2λ ) z+ (7 + 5λ) = 0
since this plane is perpendicular to the given plane x+y−3z−5 = 0, the normals of these two planes are perpendicular to each other.
Therefore, we have (1)(2 + λ) + (1)(3 + λ) + (−3)(−1 − 2λ)z = 0
which implies that λ = −1.
Thus the required equation of the plane is
(2x + 3y − z + 7)−(x + y −2z + 5) = 0
⇒ x + 2y + z + 2 = 0

We know that the vector equation of a plane passing through the line of intersection of the planes
\(\vec { r } .\vec { { n }_{ 1 } } ={ d }_{ 1 }\) and \(\vec { r } .\vec { { n }_{ 2 } } ={ d }_{ 2 }\) is given by \((\vec { r } .\vec { { n }_{ 1 } } -{ d }_{ 1 })+\lambda (\vec { r } .\vec { { n }_{ 2 } } -{ d }_{ 2 })=0\)
Substituting \(\vec { r } =x\hat { i } +y\hat { j } +z\hat { k } ,\vec { { n }_{ 1 } } =\hat { i } +\hat { j } +\hat { k } ,\vec { { n }_{ 2 } } =2\hat { i } -3\hat { j } +5\hat { k } \), \({ d }_{ 1 }=1,{ d }_{ 2 }=-2\) in the above equation, we get
(x + y + z + 1) + \(\lambda \) (2x - 3y + 5z - 2) = 0
Since this plane passes through the point (−1, 2,1) , we get λ = \(\frac{3}{5}\), and hence the required equation
of the plane is 11x−4y+20z=1 .

The Cartesian equation of the straight line joining A and B is
\(\frac { x-4 }{ 3 } =\frac { y-1 }{ 4 } =\frac { z-2 }{ 2 } \) = t (say)
Therefore, an arbitrary point on the straight line is of the form (3t + 4, 4t + 1, 2t + 2). 
To find the point of intersection of the straight line and the plane, we substitute x = 3t + 4, y = 4t + 1, z = 2t + 2 in x -y + z = 5 and we get t = 0 Therefore, the point of intersection of the straight line is (4, 1, 2) 
Now, the distance between the two points (4, 1, 2) and (5, -5, -10) is
\(\sqrt { (4-5)^{ 2 }+(1+5)^{ 2 }+(2+10)^{ 2 } } \) = \(\sqrt { 181}\) units.

We observe that the required plane is parallel to the vectors \(\vec { b } =2\hat { i } +3\hat { j } +6\hat { k } \), \(\vec { c } =\hat { i } +\hat { j } -\hat { k } \) and passing through the point(0,1,-5) with position vector \(\vec { a } \). We observe that \(\vec { b } \) is not parallel to \(\vec { c } \).
Then the vector equation of the plane in non-parametric form is given by \((\vec { r } -\vec { a } ).(\vec { b } \times \vec { c } )=0\) ..(1)
Substituting \(\vec { a } =\hat { j } -5\hat { k } \) and \(\vec { b } \times \vec { c } =\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 2 & 3 & 6 \\ 1 & 1 & -1 \end{matrix} \right| =-9\hat { i } +8\hat { j } -\hat { k } \) in equation (1), we get
\((\vec { r } (-\hat { j } -5\hat { k } )).(-9\hat { i } +8\hat { j } -\hat { k } )=0\), which implies that
\(\vec { r } (-9\hat { i } +8\hat { j } -\hat { k } )=13\)
\(\vec { r } =x\hat { i } +y\hat { j } +z\hat { k } \) is the position vector of an arbitrary point on the plane, then from the above equation, we get the Cartesian equation of the plane as -9x + 8y - z = 13 or 9x - 8y + z + 13 = 0

By definition, \(\vec { a } \times \vec { b } \) \(=\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ -2 & 3 & -2 \\ 3 & -1 & 3 \end{matrix} \right| =7\hat { i } -7\hat { k } \)
Then, \((\vec { a } \times \vec { b } )\times \vec { c } =\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 7 & 0 & -7 \\ 2 & -5 & 1 \end{matrix} \right| =-35\hat { i } -21\hat { j } -35\hat { k } \)......(1)
\(\vec { b } \times \vec { c } =\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 3 & -1 & 3 \\ 2 & -5 & 1 \end{matrix} \right| =14\hat { i } +3\hat { j } -13\hat { k } \)
\(\vec { a } (\vec { b } \times \vec { c } )=\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ -2 & 3 & -2 \\ 14 & 3 & -13 \end{matrix} \right| =-33\hat { i } -54\hat { j } -48\hat { k } \)....(2)
Therefore, equations (1) and (2) show that \((\vec { a } \times \vec { b } )\times \vec { c } \)\(​​\neq \)\((\vec { a } \times \vec { b } )\times \vec { c } \)