The given line \(\frac { x-1 }{ 6 } =\frac { y+3 }{ 2 } =\frac { z-2 }{ 3 } \) is parallel to the vector \(\overset { \rightarrow }{ b } =6\check { i } +2\check { j } +3\check { k } \)
The normal to the given plane in
\(\overset { \rightarrow }{ n } =2\check { i } +\check { j } +2\check { k } \)
\(sin\theta =\frac { \overset { \rightarrow }{ b } .\overset { \rightarrow }{ n } }{ \left| \overset { \rightarrow }{ b } \right| \left| \overset { \rightarrow }{ n } \right| } \)
\(=\frac { (6\check { i } +2\check { j } +3\check { k } )(2\check { i } -\check { j } +2k) }{ \sqrt { 39+4+9 } \sqrt { 4+1+4 } } \)
\(=\frac { 12-2+6 }{ \sqrt { 49 } \sqrt { 9 } } \)
\(\Rightarrow sin\theta =\frac { 16 }{ 7\times 3 } =\frac { 16 }{ 21 } \)
\(\therefore \theta =sin^{ -1 }\left( \frac { 16 }{ 21 } \right) \)

\(cos\theta =\left| \frac { a_{ 1 }a_{ 2 }+b_{ 1 }b_{ 2 }+c_{ 1 }c_{ 2 } }{ \sqrt { { a }_{ 1 }^{ 2 }+b_{ 1 }^{ 2 }+{ c }_{ 1 }^{ 2 } } \sqrt { { a }_{ 2 }^{ 2 }+b_{ 2 }^{ 2 }+c_{ 2 }^{ 2 } } } \right| \)
\(a_{ 1 }=7,b_{ 1 }=2,c_{ 1 }=6\)
\(a_{ 2 }=3,b_{ 2 }=-1,c_{ 2 }=-10\)
\(cos\theta =\left| \frac { 7\times 3+2\times -1+6\times -10 }{ \sqrt { 49+4+36 } \sqrt { 9+1+100 } } \right| \)
\(=\left| \frac { 21-2-60 }{ \sqrt { 89 } \sqrt { 110 } } \right| \)
\(cos\theta =\frac { 41 }{ \sqrt { 9790 } } \)
\(\Rightarrow \theta =cos^{ -1 }\left( \frac { 41 }{ \sqrt { 9790 } } \right) \)

The length of the perpendicular drawn from the origin to the plane 2x - 3y + 6z + 21 = 0
\(\frac { x }{ 5/2 } +\frac { y }{ 5 } +\frac { z }{ -5 } \)
\(=\frac { 21 }{ 7 } =3units\)

\(\check { r } +(\check { r } +2\check { j } +3\check { k } )+\lambda (\check { i } +2\check { j } -5\check { k } )\)

Let A(0, 3, 2), B(3, 5, 6)
Direction ratios of AB(a, b, c) are (3 - 0), (5 - 3), (6 - 2)
\((a_{ 1 },b_{ 1 },c_{ 1 })\) = (3, 2, 4)
Let C(1, - 1, 2), 0(3, 4, - 2)
Direction ratios of CD \((a_{ 2 },b_{ 2 },c_{ 2 })\) are (3 - 1), (4 + 1) (-2,2) \((a_{ 2 },b_{ 2 },c_{ 2 })\) are (2,56,-4)
When two lines are perpendicular if
\(a_{ 1 },a_{ 2 }+b_{ 1 },b_{ 2 },+c_{ 1 },c_{ 2 }=0\)
\(\Rightarrow 3\times 2+2\times 5+4\times -4=0\)
\(\Rightarrow 6+10-16=0\)
\(\Rightarrow 16-16=0\)
\(\therefore AB\bot to\quad CD\)

Let A be the point (5, 4, - 6)
Image A' be the point (5, 4, - 6)
\(\therefore\) A'(5, 4, - 6)
Distance between AA'
\(=\sqrt { (5-5)^{ 2 }+(4-4)^{ 2 }+(6+6)^{ 2 } } \)
\(=\sqrt { 0+0+12^{ 2 } } \)
=123 units

Getting equation as
\(\frac { x }{ 5/2 } +\frac { y }{ 5 } +\frac { z }{ -5 } =1\)
Sum of intercepts
\(\frac { 5 }{ 2 } +5-5=\frac { 5 }{ 2 } \)
Intercept magnitude is positive
So, \(\frac { 5 }{ 2 } +5+5=\frac { 25 }{ 2 } \)

The vector equation of a line parallel to Z-axis is \(\vec{m}=0 \hat{i}+0 \hat{j}+\hat{k}\). Then, the required line passes through the point A\((\alpha ,\beta ,\gamma )\), whose position vector is \(\overrightarrow{r_1}=\alpha \hat{i}+\beta \hat{j}+\gamma \hat{k}\) and is parallel to the vector \(\vec{m}=(0 \hat{i}+0 \hat{j}+\hat{k})\).
\(\therefore\)   The equation is \(\vec{r}=\overrightarrow{r_1}+\lambda \vec{m}\)
\(=(\alpha \hat{i}+\beta \hat{j}+\gamma \hat{k})+\lambda(0 \hat{i}+0 \hat{j}+\hat{k})\)
\(=(\alpha \hat{i}+\beta \hat{j}+\gamma \hat{k})+\lambda(\hat{k})\)

\( \left[ \begin{matrix} { l }_{ 4 } & { m }_{ 1 } & { n }_{ 1 } \\ { l }_{ 2 } & { m }_{ 2 } & { n }_{ 2 } \\ { l }_{ 3 } & { m }_{ 3 } & { n }_{ 3 } \end{matrix} \right] \left[ \begin{matrix} { l }_{ 1 } & { l }_{ 2 } & { l }_{ 3 } \\ { m }_{ 1 } & { m }_{ 2 } & { n }_{ 2 } \\ { n }_{ 1 } & { n }_{ 2 } & { n }_{ 3 } \end{matrix} \right] \)
\(\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right] ={ l }_{ 2 }\)
because
\({ l }_{ 1 }^{ 2 }+{ m }_{ 1 }^{ 2 }+{ n }_{ 1 }^{ 2 }=1\) for each i = 1, 2, 3
\(l_{ i }l_{ 1 }+m_{ 1 }m_{ j }+n_{ i }n_{ j }=0(i=1)\quad for\quad each\quad i,j=1,2,3\)

General point on the line is \((\lambda-2,3 \lambda+5,5 \lambda-1)\)
If it lies on the YZ - plane then \(\lambda-2=0 \Rightarrow \lambda=2\)
Point of intersection is (0, 6+ 5,10-1), i.e. )0, 11, 9.

Point through which line passes is (2, 1, - 4) and dr's: 1, 1,-1.
\(\therefore\) Cartesian equation of line is \(\frac{z-2}{1}=\frac{y-1}{-1}=\frac{z+4}{-1}\)

The given lines are:
\(\vec{r}=(\hat{i}+2 \hat{j}+\hat{k})+\lambda(\hat{i}-\hat{j}+\hat{k})\)
\(\vec{r}=2 \hat{i}-\hat{j}-\hat{k}+\mu(2 \hat{i}+\hat{j}+2 \hat{k}) \)
It is known that the shortest distance between the lines, \(\vec{r}=\vec{a}_{1}+\lambda \vec{b}_{1} \text { and } \vec{r}=\vec{a}_{2}+\mu \vec{b}_{2}, \text { is } \)
\(\text { given by, }\)
\(d=\left|\frac{\left(\vec{b}_{1} \times \vec{b}_{2}\right) \cdot\left(\vec{a}_{2}-\vec{a}_{2}\right)}{\left|\vec{b}_{1} \times \vec{b}_{2}\right|}\right|\)
Comparing the given equations, we obtain \(\)
\(\vec{a}_{1}=\hat{i}+2 \hat{j}+\hat{k} \)
\(\vec{b}_{1}=\hat{i}-\hat{j}+\hat{k} \)
\(\vec{a}_{2}=2 \hat{i}-\hat{j}-\hat{k} \)
\(\vec{b}_{2}=2 \hat{i}+\hat{j}+2 \hat{k} \)
\(\vec{a}_{2}-\vec{a}_{1}=(2 \hat{i}-\hat{j}-\hat{k})-(\hat{i}+2 \hat{j}+\hat{k})=\hat{i}-3 \hat{j}-2 \hat{k} \)
\(\vec{b}_{1} \times \vec{b}_{2}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 1 \\ 2 & 1 & 2 \end{array}\right| \)
\(\vec{b}_{1} \times \vec{b}_{2}=(-2-1) \hat{i}-(2-2) \hat{j}+(1+2) \hat{k}=-3 \hat{i}+3 \hat{k} \)
\(\Rightarrow\left|\vec{b}_{1} \times \vec{b}_{2}\right|=\sqrt{(-3)^{2}+(3)^{2}}=\sqrt{9+9}=\sqrt{18}=3 \sqrt{2} \)
Substituting all the values in equation (1), we obtain
\(d=\left|\frac{(-3 \hat{i}+3 \hat{k}) \cdot(\hat{i}-3 \hat{j}-2 \hat{k})}{3 \sqrt{2}}\right| \)
\(\Rightarrow d=\left|\frac{-3.1+3(-2)}{3 \sqrt{2}}\right| \)
\(\Rightarrow d=\left|\frac{-9}{3 \sqrt{2}}\right| \)
\(\Rightarrow d=\frac{3}{\sqrt{2}}=\frac{3 \times \sqrt{2}}{\sqrt{2} \times \sqrt{2}}=\frac{3 \sqrt{2}}{2} \)
Therefore, the shortest distance between the two lines is \(\frac{3 \sqrt{2}}{2} \text { units. } \)

The given line is 
\(\vec { r } =\left( 2\hat { i } -\hat { j } +2\hat { k } \right) +\lambda \left( 3\hat { i } +4\hat { j } +2\hat { k } \right)....(i)\)
and  the given plane is \( \vec { r } .\left( \hat { i } -\hat { j } +\hat { k } \right) =5 ....(ii)\)
Solving (i) and (ii)
\([2\hat { i } -\hat { j } +2\hat { k } +\lambda \left( 3\hat { i } +4\hat { j } +2\hat { k } \right) ].\left( \hat { i } -\hat { j } +\hat { k } \right) =5\)
\( \Rightarrow (2+3\lambda )\hat { i } +(4\lambda -1)\hat { j } +(2\lambda +2)\hat { k } ).\left( \hat { i } -\hat { j } +\hat { k } \right) =\lambda\)
\(\Rightarrow (2+3\lambda )(1)+(4\lambda -1)(-1)+(2\lambda +2)(1)=5\)
\(\Rightarrow 2+3\lambda -4\lambda +1+2\lambda +2=5\)
\(\Rightarrow \lambda =0\)
the point of intersection of(i) are (ii) is
\(\left( 2\hat { i } -\hat { j } +2\hat { k } \right) i.e.(2,-1,2).\)
The other point is \((-1,-5,-10)\)
Required distance \(=\sqrt { { \left( -1-2 \right) }^{ 2 }+{ \left( -5+1 \right) }^{ 2 }+{ \left( -10-2 \right) }^{ 2 } } =\sqrt { 9+16+144 } =13\)

Here, \(x_{1}=a-d x_{2}=b-c \)
\(y_{1}=a , y_{2}=b \)
\(z_{1}=a+d , z_{2}=b+c \)
\(a_{1}=\alpha-\delta , a_{2}=\beta-\gamma \)
\(b_{1}=\alpha , b_{2}=\beta \)
\(c_{1}=\alpha+\delta , c_{2}=\beta+\gamma \)
Now consider the determinant
\(\left|\begin{array}{ccc} x_{2}-x_{1} & y_{2}-y_{1} & z_{2}-z_{1} \\ a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \end{array}\right|=\left|\begin{array}{ccc} b-c-a+d & b-a & b+c-a-d \\ \alpha-\delta & \alpha & \alpha+\delta \\ \beta-\gamma & \beta & \beta+\gamma \end{array}\right|\)
Adding third column to the first column, we get
\(2\left|\begin{array}{ccc} b-a & b-a & b+c-a-d \\ \alpha & \alpha & \alpha+\delta \\ \beta & \beta & \beta+\gamma \end{array}\right|=0\)
Since the first and second columns are identical. Hence, the given two lines are coplanar.

Given, lines are
\(\begin{aligned} \frac{x+1}{3}=\frac{y+3}{5}=\frac{z+5}{7}=\lambda \end{aligned}\)   (let)           ...(i)
and \(\begin{aligned} \frac{x-2}{1}=\frac{y-4}{3}=\frac{z-6}{5}=\mu \end{aligned}\) (let)      ...(ii)
Then, any point on line (i) is of the form
\(P(3 \lambda-1,5 \lambda-3,7 \lambda-5)\)                       ...(iii)
and any point on line (ii) is of the form
\(Q(\mu+2,3 \mu+4,5 \mu+6)\)                         ...(iv)
If lines (i) and (ii) intersect, then these points must coincide for some \(\lambda\) and \(\mu\).
Consider,  \(\begin{aligned} 3 \lambda-1 & =\mu+2 \end{aligned}\)
             \(\begin{aligned} 5 \lambda-3 & =3 \mu+4 \end{aligned}\)
and           \(\begin{aligned} 7 \lambda-5 & =5 \mu+6 \end{aligned}\)
\(\Rightarrow\)           \(\begin{aligned} 3 \lambda-\mu & =3 \end{aligned}\)                              ...(v)
                   \(\begin{aligned} 5 \lambda-3 \mu & =7 \end{aligned}\)                           ...(vi)
and              \(\begin{aligned} 7 \lambda-5 \mu & =11 \end{aligned}\)                        ...(vii)
On multiplying Eq. (v) by 3 and then subtracting Eq. (vi), we get
\(9 \lambda-3 \mu-5 \lambda+3 \mu=9-7 \)
\(\Rightarrow \quad 4 \lambda=2\)
\(\Rightarrow \quad \lambda=\frac{1}{2} \)
On putting the value of \(\lambda\) in Eq. (v),we get
\(3 \times \frac{1}{2}-\mu=3 \Rightarrow \frac{3}{2}-\mu=3\)
\(\Rightarrow \quad \mu=-\frac{3}{2}\)
On putting the values of \(\lambda\) and \(\mu\) in Eq. (vii), we get
\(\begin{aligned} 7 \times \frac{1}{2}-5\left(-\frac{3}{2}\right) & =11 \end{aligned}\)
\(\begin{aligned} \Rightarrow \quad \frac{7}{2}+\frac{15}{2} & =11 \Rightarrow \frac{22}{2}=11 \end{aligned}\)
\(\Rightarrow\) 11 = 11, which is true.
Hence, lines (i) and (ii) intersect and their point of intersection is
\(P\left(3 \times \frac{1}{2}-1,5 \times \frac{1}{2}-3,7 \times \frac{1}{2}-5\right)\)
\(\text { [put } \lambda=\frac{1}{2} \text { in Eq. (iii)] }\)
i.e.  \(P\left(\frac{1}{2},-\frac{1}{2},-\frac{3}{2}\right)\)

\(\sin { \theta } =\frac { 2\times 10+3\times 2+6\times (-11) }{ \sqrt { 4+9+36 } \sqrt { 100+4+121 } } =\frac { 20+6-66 }{ 7\times 15 } =\frac { -8 }{ 21 }\)
\( \Rightarrow \theta ={ sin }^{ -1 }\left( \frac { -8 }{ 21 } \right) \)units

\(\vec { n } =\vec { { b }_{ 1 } } \times \vec { { b }_{ 2 } } =\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 1 & -1 & 2 \\ 2 & -1 & -3 \end{matrix} \right| =5\hat { i } +7\hat { j } +\hat { k } \)
The equation of the plane is \(\vec { r } .\vec { n } =\left( 5\hat { i } +7\hat { j } +\hat { k } \right) .\left( 4\hat { i } -3\hat { j } +2\hat { k } \right) ,\)
i.e., \(\vec { r } =\left( 5\hat { i } +7\hat { j } +\hat { k } \right) =1\)
The position vector of any on the given line is \(\left( 1+\lambda \right) \hat { i } +\left( 2+3\lambda \right) \hat { j } +\left( -1-9\lambda \right) \hat { k } \)
We have \(\left( 1+\lambda \right){ 5} +\left( 2+3\lambda \right) { 7 } +\left( -1-9\lambda \right) {1 } =1\)
\(\lambda =-1\)
The position vector of the required point is \(-\hat { j } +8\hat { k } \).

Let equation of plane through (1, 2, -4) be a(x-1) + b(y-2) + c(z+4) = 0.
The plane is parallel to the given lines
\(\therefore\)  2a + 3b + 6c = 0; a + b - c = 0
Solving: \(\frac { a }{ -9 } =\frac { b }{ 8 } =\frac { c }{ -1 } =k\left( say \right) \)
\(\therefore\) a = -9k, b = 8k, c = -k
From (i), -9k(x-1) + 8k(y-2) - k(z+4) = 0
\(\therefore\) Equation of plane in cartesian form is 9x - 8y + z+11 = 0
Vector form of plane is: \(\Rightarrow \vec { r } .\left( 9\hat { i } -8\hat { j } +\hat { k } \right) =-11\)
Distance of (9, -8, -10) from the plane = \(\left| \frac { 9.9-8\left( -8 \right) +1\left( -10 \right) +11 }{ \sqrt { 81+64+1 } } \right| =\sqrt { 146 } \)

Equation of plane through the given lines is
\(\left\{ \vec { r } .\left( 2\hat { i } +2\hat { j } -3\hat { k } \right) -7 \right\} +\lambda \left\{ \left( 2\hat { i } +5\hat { j } +3\hat { k } \right) -9 \right\} =0......\left( i \right) \)
\(\Rightarrow \vec { r } .\left\{ \left( 2+2\lambda \right) \hat { i } +\left( 2+5\lambda \right) \hat { j } +\left( -3+3\lambda \right) \hat { k } \right\} =\left( 7+9\lambda \right) .\left( ii \right) \)
Here, x intercept = z intercept 
\(\therefore \frac { 7+9\lambda }{ 2+2\lambda } =\frac { 7+9\lambda }{ -3+3\lambda } \Rightarrow \lambda =5\)
\(\therefore\)  Equation of plane in vector form is obtained by putting the value of in equation (ii).
i.e., \(\vec { r } .\left( 12\hat { i } +27\hat { j } +12\hat { k } \right) =52\) and equation of plane in cartesian form is
given as \(\left( x\hat { i } +y\hat { j } +z\hat { k } \right) .\left( 12\hat { i } +27\hat { j } +12\hat { k } \right) =52\)
i.e., 12x + 27y + 12z - 52 = 0