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)

If the position vector of the given point is \(\vec { a } =4\hat { i } +2\hat { j } -3\hat { k } \) and \(\vec { n } =2\hat { i } -\hat { j } +\hat { k } \), then the equation of the plane passing through a point and normal to a vector is given by \((\vec { r } -\vec { a } ).\vec { n } =0\) or \(\vec { r } .\vec { n } =\vec { a } .\vec { n } \)
Substituting \(\vec { a } =4\hat { i } +2\hat { j } -3\hat { k } \) and \(\vec { n } =2\hat { i } -\hat { j } +\hat { k } \) in the above equation, we get
\(\vec { r } .(2\hat { i } -\hat { j } +\hat { k } )=(4\hat { i } +2\hat { j } -3\hat { k } ).(2\hat { i } -\hat { j } +\hat { k } )\)
Thus, the required vector equation of the plane is \(\vec { r } .(2\hat { i } -\hat { j } +\hat { k } )\)= 3. If \(\vec { r } =x\hat { i } +y\hat { j } +z\hat { k } \) then
we get the Cartesian equation of the plane 2x − y + z = 3.

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 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.

Since the vectors \(\vec { a } ,\vec { b } ,\vec { c } \) are coplanar, we have \([\vec { a } ,\vec { b } ,\vec { c } ]\) = 0 Using the properties of the scalar triple product, we get
\([\vec { a } +\vec { b } ,\vec { b } +\vec { c } ,\vec { c } +\vec { a } ]=[\vec { a } ,\vec { b } +\vec { c } ,\vec { c}+\vec {a } ]+[\vec { b } ,\vec { b } +\vec { c } ,\vec { c } +\vec { a } ]\)
\(=[\vec { a } ,\vec { b } ,\vec { c } +\vec { a } ]+[\vec { a } ,\vec { c } ,\vec { c } +\vec { a } ]+[\vec { b } ,\vec { b } ,\vec { c } +\vec { a } ]+[\vec { b } ,\vec { c } ,\vec { c } +\vec { a } ]\)
\(=[\vec { a } ,\vec { b } ,\vec { c } ]+[\vec { a } ,\vec { b } ,\vec { a } ]+[\vec { a } ,\vec { c } ,\vec { c } ]+[\vec { a } ,\vec { c } ,\vec { a } ]+[\vec { b } ,\vec { b } ,\vec { c } ]+[\vec { b } ,\vec { b } ,\vec { a } ]+[\vec { b } ,\vec { c } ,\vec { c } ]+[\vec { b } ,\vec { c } ,\vec { a } ]\)
\(=[\vec { a } ,\vec { b } ,\vec { c } ]+[\vec { a } ,\vec { b } ,\vec { c } ]=2[a,b,c]=0\)
Hence the vectors \(\vec { a } +\vec { b } ,\vec { b } +\vec { c } ,\vec { c } +\vec { a } \) are coplanar.

Let the forces be \(\vec { { F }_{ 1 } } \) and \(\vec { { F }_{ 2 } } \) and \(\vec { d } \) be the displacement vector
∴ Resultant force \(\vec { F } =\vec { { F }_{ 1 } } +\vec { F_{ 2 } } \)
= \((8\hat { i } +2\hat { j } -6\hat { k } )+(6\hat { i } +2\hat { j } -2\hat { k } )\)
\(\vec { F } =14\hat { i } +4\hat { j } -8\hat { k } \)
\(\vec { d } \)= Displacement to the point - displacement from the point
= \((5\hat { i } +4\hat { j } +\hat { k } )-(\hat { i } +2\hat { j } +\hat { k } )\)
= \(4\hat { i } +2\hat { j } -2\hat { k } \)
Work done (w) = \(\vec { F } .\vec { d } \)
= \((14\hat { i } +4\hat { j } -8\hat { k } ).(4\hat { i } +2\hat { j } -2\hat { k } )\)
= 56 + 8 + 16
= 80 units.

Resultant of the given forces is \(\vec { F } \) = \((\hat { 3i } -\hat { 2j } +\hat { 2k } )\) + \((\hat { 2i } +\hat { j } -\hat { k } )\) = \(​​\hat { 5i } -\hat { j } +\hat { k } \)
The displacement of the particle is given by
\(\vec { d } \) =  \((\hat { 4i } -\hat { j } +\hat { \lambda k } )-(\hat { i } +3\hat { j } -\hat { k } )\) =  \((3\hat { i } -\hat { 4j } +(\lambda +1)\hat { k } )\)
As the work done by the forces is 16 units, we have
\(\vec { F } \).\(\vec { d } \) = 16
That is \((\hat { 5i } -\hat { j } +\hat { k } ).(3\hat { i } -\hat { 4j } +(\lambda +1))\hat { k } \)  = 16 ⇒ λ + 20 = 16
So, λ = - 4

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.

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.

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.

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.

By the defination of scalar triple product of three vectors,
We find, \(\hat { a } .(\hat { b } \times \hat { c } )\) = \(\left| \begin{matrix} -3 & -1 & 5 \\ 1 & -2 & 1 \\ 0 & 4 & -5 \end{matrix} \right| =-3\)

The normal vector to the planes
x + 2y + 3z - 7 = 0, 2x - 3y + 4z = 0 are 
\(\overset { \rightarrow }{ b } =\overset { \wedge }{ i } +2\overset { \wedge }{ j } +3\overset { \wedge }{ k } \) and \(\overset { \rightarrow }{ c } =2\overset { \wedge }{ i } -3\overset { \wedge }{ j } +4\overset { \wedge }{ k } \)
∴ The required planes passes through the point \(\overset { \rightarrow }{ a } =\overset { \wedge }{ i } +\overset { \wedge }{ j } +\overset { \wedge }{ k } \) and parallel to two vector 5 namely \(\overset { \rightarrow }{ b } \) and \(\overset { \rightarrow }{ c } \) 
∴ The Parametric form of vectors equation of the plans is \(\overset { \rightarrow }{ r } =\overset { \rightarrow }{ a } +s\overset { \rightarrow }{ b } +t\overset { \rightarrow }{ c } \) s, t ∈ R
\(\overset { \rightarrow }{ r } =\left( \overset { \rightarrow }{ i } +\overset { \rightarrow }{ j } -\overset { \rightarrow }{ k } \right) +s\left( \overset { \rightarrow }{ i } +2\overset { \rightarrow }{ j } +3\overset { \rightarrow }{ k } \right) +t\left( 2\overset { \rightarrow }{ i } -3\overset { \rightarrow }{ j } +4\overset { \rightarrow }{ k } \right) ,\)
Cartesian equation is \(\left| \begin{matrix} x-{ x }_{ 1 } \\ { b }_{ 1 } \\ { c }_{ 1 } \end{matrix}\begin{matrix} y-{ { y }_{ 1 } } \\ { b }_{ 2 } \\ { c }_{ 2 } \end{matrix}\begin{matrix} z-{ { z }_{ 1 } } \\ { b }_{ 3 } \\ { c }_{ 3 } \end{matrix} \right| =0\)
\(\Rightarrow \left| \begin{matrix} x-1 \\ 1 \\ 2 \end{matrix}\begin{matrix} y-1 \\ 2 \\ -3 \end{matrix}\begin{matrix} z+1 \\ 3 \\ 4 \end{matrix} \right| =0\)
⇒ (x - 1) (8 + 9) - (y - 1)(4 - 6) + (z + 1)(-3 -4) = 0
⇒ 17 (x - 1) +2 (y - 1) -7 (z + 1) = 0
⇒ 17x - 17 + 2y - 2 - 7z - 7 = 0
⇒ 17x + 2y - 7z - 26 = 0

From the line  \(\frac { x-3 }{ 3 } =\frac { y-8 }{ -1 } =\frac { z-3 }{ 1 } \), we get
\(\overset { \rightarrow }{ a } =3\overset { \wedge }{ i } +8\overset { \wedge }{ j } +3\overset { \wedge }{ k } \), \(\overset { \rightarrow }{ b } =3\overset { \wedge }{ i } -\overset { \wedge }{ j } +\overset { \wedge }{ k } \)
From the line \(\frac { x+3 }{ -3 } =\frac { y+7 }{ 2 } =\frac { z-6 }{ 4 } \) we get
\(\overset { \rightarrow }{ c } =-3\overset { \wedge }{ i } -7\overset { \wedge }{ j } +6\overset { \wedge }{ k } \) and \(\overset { \rightarrow }{ d } =-3\overset { \wedge }{ i } +2\overset { \wedge }{ j } +4\overset { \wedge }{ k } \)
Since the given lines are not parallel, the shortest distance between the line is 
\(d=\left| \frac { \left( \overset { \rightarrow }{ c } -\overset { \rightarrow }{ a } \right) .\left( \overset { \rightarrow }{ b } \times \overset { \rightarrow }{ d } \right) }{ \left| \overset { \rightarrow }{ b } \times \overset { \rightarrow }{ d } \right| } \right| \)
\(\overset { \rightarrow }{ c } -\overset { \rightarrow }{ a } =-6\overset { \wedge }{ i } -15\overset { \wedge }{ j } +3\overset { \wedge }{ k } \)
\(\overset { \rightarrow }{ b } \times \overset { \rightarrow }{ d } =\left| \begin{matrix} \overset { \wedge }{ i } \\ 3 \\ -3 \end{matrix}\begin{matrix} \overset { \wedge }{ j } \\ -1 \\ 2 \end{matrix}\begin{matrix} \overset { \wedge }{ k } \\ 1 \\ 4 \end{matrix} \right| \)
\(=\overset { \wedge }{ i } (-4-2)-\overset { \wedge }{ j } (12+3)+\overset { \wedge }{ k } (6-3)\\ \)
\(=-6\overset { \wedge }{ i } -15\overset { \wedge }{ j } +3\overset { \wedge }{ k } \)
\(\therefore \left| \overset { \rightarrow }{ b } \times \overset { \rightarrow }{ d } \right| =\sqrt { 36+225+9 } \)
\(=\sqrt { 270 } \)
\(\therefore \left( \overset { \rightarrow }{ c } -\overset { \rightarrow }{ a } \right) .\left( \overset { \rightarrow }{ b } \times \overset { \rightarrow }{ d } \right) =\left( -6\overset { \wedge }{ i } -15\overset { \wedge }{ j } +3\overset { \wedge }{ k } \right) .\left( -6\overset { \wedge }{ i } -15\overset { \wedge }{ j } +3\overset { \wedge }{ k } \right) \)
= -6 (-6) + 15 _(15) + 3(3)
= 270 ≠ 0
Since the given lines are neither intersecting, nor parallel they are skew lines
\(\therefore d=\frac { 270 }{ \sqrt { 270 } } =\sqrt { 270 } units\)

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.

Comparing the given equation \(\vec { r } =(\hat { i } -4\hat { j } +3\hat { k } )+t(2\hat { i } +3\hat { j } +\hat { k } )\) with \(\vec { r } =\hat { a } +t\hat { b } \),
we get \(\vec { a } =\hat { i } -4\hat { j } +3\hat { k } \) and \(\vec { b } =2\hat { i } +3\hat { j } +\hat { k } \). We denote the given point (-1, 2, 3) by D and the point (1, -4, 3) on the straight line by F.
If F is the foot of the perpendicular from to the straight line, then F is of the form (2t + 1, 3t - 4, t+3)  and \(\vec { DF } =\vec { OF } -\vec { OD } =(2t+2)\hat { i } +(3t-6)\hat { j } +t\hat { k } \)

Since \(\vec { b } \) is perpendicular to \(\vec { DF } \), we have
\(\vec { b } .\vec { DF } \) = 0 ⇒ 2(2t + 2)+ 3(3t - 6) + 1(t) ⇒ t= 1
Therefore, the coordinate of F is (3, -1, 4)
Now, the perpendicular distance from the given point to the given line is
\(DF=\left| \vec { DF } \right| =\sqrt { { 4 }^{ 2 }+(-{ 3) }^{ 2 }+{ 1 }^{ 2 } } =\sqrt { 26 } \)

Let A = (6, -7, 0), B = (16, -19, -4), C = (0, 3, -6), D = (2, -5, 10). To show that the four points A, B, C, D lie on a plane,
we have to prove that the three vectors \(\vec { AB } ,\vec { AC } ,\vec { AD } \)  are coplanar.
Now, \(\vec { AB } =\vec { OB } -\vec { OA } =(16\hat { i } -19\hat { j } -4\hat { k } )-(6\hat { i } -7\hat { j } )=10\hat { i } -12\hat { j } -4\hat { k } \)
\(\vec { AC } =\vec { OC } -\vec { OA } =-6\hat { i } +10\hat { j } -6\hat { k } \) and \(\vec { AC } =\vec { OC } -\vec { OA } =-6\hat { i } +10\hat { j } -6\hat { k } \)
We have \([\vec { AB } ,\vec { AC } ,\vec { AD } ]\) = \(\left| \begin{matrix} 10 & -12 & -4 \\ -6 & 10 & -6 \\ -4 & 2 & 10 \end{matrix} \right| \) = 0
Therefore, the three vectors \(\vec { AB } ,\vec { AC } ,\vec { AD } \) are coplanar and hence the four points A, B, C and D lie on a plane