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 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.

The required plane is parallel to the given line and so it is parallel to the vector \(\vec { c } =\hat { i } +\hat { j } -\hat { k } \) and the plane passes through the points \(\vec { a } =-\hat { i } +2\hat { j } ,\vec { b } =2\hat { i } +2\hat { j } -\hat { k } \)
(i) vector equation of the plane in parametric form is \(\vec { r } =\vec { a } +s(\vec { b } -\vec { a } )+t\vec { c } \), where s, t ∈ R
which implies that \(\vec { r } =(-\hat { i } +2\hat { j } )+s(3\hat { i } -\hat { k } )+t(\hat { i } +\hat { j } -\hat { k } )\), where s, t ∈ R
(ii) vector equation of the plane in non-parametric form is \((\vec { r } -\vec { a } ).(\vec { b } -\vec { a } )\times \vec { c } )\) = 0
Now, \((\vec { b } -\vec { a } )\times \vec { c } =\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 3 & 0 & -1 \\ 1 & -1 & -1 \end{matrix} \right| =\hat { i } +2\hat { j } +3\hat { k } \)
we have \((\vec { r } -(-\hat { i } +2\hat { j } ).(\hat { i } +2\hat { j } +3\hat { k } )\) = 0 ⇒ \(\vec { r } .(\hat { i } +2\hat { j } +3\hat { k } )\) = 3
If  \(\vec { r } .(\hat { i } +2\hat { j } +3\hat { k } )\) 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 x + 2y + 3z = 3

Let \(\hat { a } =\vec { OA } \) and \(\hat { b } =\vec { OB } \) be the unit vectors making angles α and β respectively, with positive x -axis, where A and B are as shown in the diagram.
Then, we get
\(\hat { a } \) = cosα\(​​\hat{i}\)+ sinα\(\hat {j } \) and \(\hat{b}\)ˆ = cosβ\(​​\hat{i}\)+ sin β \(\hat {j } \),

The angle between \(\hat { a } \) and \(\hat{b}\) is α −β and, the vectors \(\hat{b}\), \(\hat { a } \), \(\hat{k}\) form a right-handed system.
Hence, we get \(\hat { b } \times \hat { a } =\left| \hat { b } \right| \left| \hat { a } \right| \)sin(α −β )\(\hat {k}\) = sin(α −β)\(\hat {k}\)
On the other hand, \(\hat { b } \times \hat { a } =\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ cos\beta & sin\beta & 0 \\ sin\alpha & sin\alpha & 0 \end{matrix} \right| \) = (sin α cos β -cos α sin β)\(\hat { k } \)
Hence, by equations (1) and (2), we get
sin(α −β ) = sin α cos β − cos α sin β

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.

Let \(\vec { a } =-6\hat { i } +14\hat { j } +10\hat { k } \), \(\vec { b } =14\hat { i } -10\hat { j } -6\hat { k } \) and \(\vec { c } =2\hat { i } +4\hat { j } -2\hat { k } \)
Volume of the parallelepiped having \(\vec { a } ,\vec { b } \) and \(\vec { c } \) as its co-terminus edges is \(\vec { a } .(\vec { b } \times \vec { c } )\).
∴ \(\vec { a } .(\vec { b } \times \vec { c } )=\left| \begin{matrix} -6 & 14 & 10 \\ 14 & -10 & -6 \\ 2 & 4 & -2 \end{matrix} \right| \)
= \(-6\left| \begin{matrix} -10 & -6 \\ 4 & -2 \end{matrix} \right| -14\left| \begin{matrix} 14 & -6 \\ 2 & -2 \end{matrix} \right| +10\left| \begin{matrix} 14 & -10 \\ 2 & 4 \end{matrix} \right| \)
= -6(20 + 24) - 14(-28 + 12) + 10(56 + 20)
= -6(44) -14(-16) + 10(76)
= -264 + 224 + 760 = 720.
∴ Volume of the required parallelepiped = 720 cubic units.

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

From the line \(\frac { x-1 }{ 3 } =\frac { y+1 }{ 2 } =\frac { z-1 }{ 5 } \)
(x₁, y₁, z₁) is (1, -1, 1)
(l₁, m₁, n₁) is (3, 2, 5)
From the line \(\frac { x+2 }{ 4 } =\frac { y-1 }{ 3 } =\frac { z+1 }{ -2 } \)
we get, (x₂, y₂, z₂) is (-2, 1, -1)
(l₂, m₂, n₂) is 4, 3, -2
The Condition for intersecting lines is 
\(\left| \begin{matrix} { x }_{ 2 }-{ x }_{ 1 } \\ { l }_{ 1 } \\ { l }_{ 2 } \end{matrix}\begin{matrix} { y }_{ 2 }-{ y }_{ 1 } \\ { m }_{ 1 } \\ { m }_{ 2 } \end{matrix}\begin{matrix} { z }_{ 2 }-{ z }_{ 1 } \\ { n }_{ 1 } \\ { n }_{ 2 } \end{matrix} \right| =0\)
\(\Rightarrow \left| \begin{matrix} -2-1 \\ 3 \\ 4 \end{matrix}\begin{matrix} 1+1 \\ 2 \\ 3 \end{matrix}\begin{matrix} -1-1 \\ 5 \\ -2 \end{matrix} \right| \)
\(\Rightarrow \left| \begin{matrix} -3 \\ 3 \\ 4 \end{matrix}\begin{matrix} 2 \\ 2 \\ 3 \end{matrix}\begin{matrix} -2 \\ 5 \\ -2 \end{matrix} \right| \)
= -3 (-4 -15) -2 (-6 -20) -2 (9 - 8)
= -3(-19) - 2(-26) -2 (1)
= 57 + 52 - 2 = 57 + 50
= 107 ≠ 0
Hence the given lines do not intersect

The equation of the requir d plane through the intersection of the given plane is
(2x - 3y + z - 4) + λ (x- y + z + 1) = 0      (1)
⇒ (2 + λ) x - (3 + λ) y + (1 + λ) z - 4 + λ = 0
Since this plane perpendicular to x + 2y - 3z + 6 - 0,
We have 1(2 + λ) - 2(3 + λ) - 3(1 + λ)= 0
⇒ 2 + λ - 6 - 2λ - 3 - 3λ = 0
⇒ -4λ - y = 0
⇒ \(\lambda =\frac { 7 }{ 4 } \)
Substituting \(\lambda =\frac { 7 }{ 4 } \) in (1) we get,
(2x - 3y + z - 4) - \(\frac { -7 }{ 4 } \) (x - y + z + 1) = 0
⇒ 4 (2x - 3y + z -4) -7 (x - y + z + 1) = 0
⇒ x - 5y - 3z - 23 = 0 which is the equation of the required plane.

Let \(\overset { \rightarrow }{ a } =\overset { \wedge }{ 3i } -5\overset { \wedge }{ k } ,\overset { \rightarrow }{ b } =2\overset { \wedge }{ i } +7\overset { \wedge }{ j } \) and \(\overset { \rightarrow }{ c } =\overset { \wedge }{ i } +\overset { \wedge }{ j } +\overset { \wedge }{ k } \)
Let the required vector be \(\overset { \rightarrow }{ r } =x\overset { \wedge }{ i } +y\overset { \wedge }{ j } +z\overset { \wedge }{ k } \)
Given \(\overset { \rightarrow }{ r } .\overset { \rightarrow }{ a } =-1\)
\(\Rightarrow \left( x\overset { \wedge }{ i } +y\overset { \wedge }{ j } +z\overset { \wedge }{ k } \right) .\left( \overset { \wedge }{ 3i } -5\overset { \wedge }{ k } \right) =-1\)
⇒ 3x - 5z = -1     (1)
\(\overset { \rightarrow }{ r } .\overset { \rightarrow }{ b } =6\)
\(\Rightarrow \left( x\overset { \wedge }{ i } +y\overset { \wedge }{ j } +z\overset { \wedge }{ k } \right) .\left( 2\overset { \wedge }{ i } +7\overset { \wedge }{ j } \right) \)= 2x + 7y = 6       (2)
\(\overset { \rightarrow }{ r } .\overset { \rightarrow }{ i } =5\)
\(\Rightarrow \left( x\overset { \wedge }{ i } +y\overset { \wedge }{ j } +z\overset { \wedge }{ k } \right) .\left( \overset { \wedge }{ i } +\overset { \wedge }{ j } +\overset { \wedge }{ k } \right) \)= x + y + z = 5   (3)
Solving (1), (2) and (3) we get
x = 3, y = 0 and z = 2.
\(\therefore \overset { \rightarrow }{ r } =\overset { \wedge }{ 3i } +2\overset { \wedge }{ k } \)

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)

\(\hat { d } =\frac { 3\hat { i } -4\hat { j } +5\hat { k } }{ \sqrt { { 3 }^{ 2 }+(-4)+{ 5 }^{ 2 } } } =\frac { 3\hat { i } -4\hat { j } +5\hat { k } }{ \sqrt { 9+16+25 } } \)
\(=\frac { 3\hat { i } -4\hat { j } +5\hat { k } }{ \sqrt { 50 } } =\frac { 3\hat { i } -4\hat { j } +5\hat { k } }{ 5\sqrt { 2 } } \)
[∵ 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
\(\vec { r } .\left( \frac { 3\hat { i } -4\hat { j } +5\hat { k } }{ 5\sqrt { 2 } } \right) =7\)

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
\((\vec { a } \times \vec { b } )\times (\vec { a } \times \vec { c } )=((\vec { a } \times \vec { b } ).\vec { c } )\vec { a } -((\vec { a } \times \vec { b } ).\vec { a } )\vec { c } =(\vec { a } .\vec { b } \times \vec { c } ))\vec { a } \)