Given planes are \(\vec { r } .(\hat { i } +\hat { j } -2\hat { k } )\) and 
\(2x-2y+z=2\Rightarrow \vec { r } .\left( 2\hat { i } -2\hat { j } +\hat { k } \right) =3\)
\(\therefore { \vec { n } }_{ 1 }=\hat { i } +\hat { j } -2\hat { k } \) and \({ \vec { n } }_{ 2 }=2\hat { i } -2\hat { j } +\hat { k } \)
Angle between the plane is'
\(cos\theta =\frac { { \vec { n } }_{ 1 }.{ \vec { n } }_{ 2 } }{ \left| { \vec { n } }_{ 1 } \right| \left| { { \vec { n } }_{ 2 } } \right| } =\frac { \left| 1(2)+1(-2)-2(1) \right| }{ \sqrt { { 1 }^{ 2 }+{ 1 }^{ 2 }+\left( -2 \right) ^{ 2 }.\sqrt { { 2 }^{ 2 }+\left( -2 \right) ^{ 2 }+{ 1 }^{ 2 } } } } \)
= \(\frac { \left| -2 \right| }{ \sqrt { 6 } .\sqrt { 9 } } =\frac { 2 }{ \sqrt { 6 } (3) } =\frac { 2 }{ 3\sqrt { 6 } } \)
\(\theta ={ { cos }^{ -1 }\left( \frac { 2 }{ 3\sqrt { 6 } } \right) }\)

Equation of given plane is
\(\vec { r } .\left( 6\hat { i } +3\hat { j } +2\hat { k } \right) =8\)
\(\therefore { \vec { n } }_{ 1 }=6\hat { i } +3\hat { j } +2\hat { k } \)
and the line is \(\vec { r } =\left( 2\hat { i } -\hat { j } +\hat { k } \right) +t\left( \hat { i } +2\hat { j } -2\hat { k } \right) \)
\(\therefore \vec { b } =\hat { i } +2\hat { j } -2\hat { k } \)
Angle between a line and a plane is
\(sin\theta =\frac { \left| \vec { b } .\vec { n } \right| }{ \left| \vec { b } \right| \left| \vec { n } \right| } =\cfrac { \left| 6(1)+3(2)+2(-2) \right| }{ \sqrt { { 1 }^{ 2 }+{ 2 }^{ 2 }+\left( -2 \right) ^{ 2 }.\sqrt { { 6 }^{ 2 }+{ 3 }^{ 2 }+{ 2 }^{ 2 } } } } \)
= \(\frac { 6+6-4 }{ \sqrt { 9 } .\sqrt { 36+9+4 } } =\frac { 8 }{ \sqrt { 9 } .\sqrt { 49 } } =\frac { 8 }{ 3\left( 7 \right) } =\frac { 8 }{ 21 } \)
\(\therefore \theta ={ sin }^{ -1 }\left( \cfrac { 8 }{ 21 } \right) \)

\(\vec { a } =-\hat { i } +\hat { j } +2\hat { k } \)
Given \(|\vec { n } |=3\sqrt { 3 } \) and let \(\alpha\) be the angle made by the normal with the co-ordinate axes.
\(\therefore { cos }^{ 2 }\alpha +{ cos }^{ 2 }\alpha =1\Rightarrow 3{ cos }^{ 2 }\alpha =1\)
\(\Rightarrow { cos }^{ 2 }\alpha =\frac { 1 }{ 3 } { cos }\alpha =\frac { 1 }{ \sqrt { 3 } } \)
\(\therefore \vec { n } =3\sqrt { 3 } \left( \frac { 1 }{ \sqrt { 3 } } \hat { i } +\frac { 1 }{ \sqrt { 3 } } \hat { j } +\frac { 1 }{ \sqrt { 3 } } \hat { k } \right) =3\hat { i } +3\hat { j } +3\hat { k } \)
∴ The equation of the required plane is
\(\vec { r } .\vec { n } =\vec { a } .\vec { n } \Rightarrow \vec { r } .(3\hat { i } +3\hat { j } +3\hat { k } )\)
\(=(-\hat { i } +\hat { j } +2\hat { k } ).(3\hat { i } +3\hat { j } +3\hat { k } )\)
\(\Rightarrow \vec { r } .(3\hat { i } +3\hat { j } +3\hat { k } )=-3+3+6=6\)
\(\Rightarrow \vec { r } .\frac { (3\hat { i } +3\hat { j } +3\hat { k } ) }{ 2 } =\frac { 6 }{ 3 } =2\)
\(\hat { r } .(\hat { i } +\hat { j } +\hat { k } )\)
Let \(\vec { r } =x\hat { i } +y\hat { j } +z\hat { k } \)
\(\Rightarrow (x\hat { i } +y\hat { j } +z\hat { k } ).(\hat { i } +\hat { j } +\hat { k } )=2\Rightarrow x+y+z=2\) which is the equation of the required plane.

Given \(\vec { a } \) = \(2\hat { i } +6\hat { j } +3\hat { k } \)
\(\vec { n } \) = \(\hat { i } +3\hat { j } +5\hat { k } \)
Vector form of the evaluation of the plane passing through one point (\(\vec { a } \)) and normal to a vector (\(\vec { n } \)) is
\(\vec { r } .\vec { n } =\vec { a } .\vec { n }\)
\(=\vec { r } .(\hat { i } +3\hat { j } +5\hat { k } )=(2\hat { i } +6\hat { j } +3\hat { k } ).(\hat { i } +3\hat { j } +5\hat { k } )\)
= 2 + 18 + 15
\(\Rightarrow \vec { r } .(\hat { i } +3\hat { j } +5\hat { k } )=35\)
Its Cartesian equation will be
\(a(x-{ x }_{ 1 })+b(y-{ y }_{ 1 })+c(z-{ z }_{ 1 })=0\)
1(x-2)+3(y-6)+5(z-3) = 0
[\(\because ({ x }_{ 1 },{ y }_{ 1 },{ z }_{ 1 }\) is (2, 6, 3) and a, b, c = 1, 3, 5]
\(\Rightarrow\) x-2 + 3y-18 + 5z-15 = 0
\(\Rightarrow\) x + 3y + 5z - 35 = 0
\(\Rightarrow\) x + 3y + 5z = 35

The equation of the plane having intercepts a, b, c on the x, y, z axes respectively is \(\frac { x }{ a } +\frac { y }{ b } +\frac { z }{ c } =1\). 
Since the sum of the reciprocals of the intercepts on the coordinate axes is a constant, we have \(\frac { 1 }{ a } +\frac { 1 }{ b } +\frac { 1 }{ c } =k\),  where k is a constant, and which can be written as \(\frac { 1 }{ a } \left( \frac { 1 }{ k } \right) +\frac { 1 }{ b } \left( \frac { 1 }{ k } \right) +\frac { 1 }{ c } \left( \frac { 1 }{ k } \right) =1\)
This shows that the plane \(\frac { x }{ a } +\frac { y }{ b } +\frac { z }{ c } =1\) passes through the fixed point \(\left( \frac { 1 }{ k } ,\frac { 1 }{ k } ,\frac { 1 }{ k } \right) \)

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 \(\vec { a } =-2\hat { i } +3\hat { j } +4\hat { k } \) and \(\vec { b } =-4\hat { i } +5\hat { j } -6\hat { k } \)
The parametric form of vector equation of a straight line passing through a point \((\vec { b } )\) and parallel to is \(\vec { b } \)is
\(\vec { r } =\vec { a } +t\vec { b } \) where \(t\in R\)
∴ \(\vec { r } =-2\hat { i } +3\hat { j } +4\hat { k } +t(-4\hat { i } +5\hat { j } -6\hat { k } ),t\in R\)
Its Cartesian equation is
\(\frac { x-{ x }_{ 1 } }{ { b }_{ 1 } } =\frac { y-{ y }_{ 1 } }{ { b }_{ 2 } } =\frac { z-{ z }_{ 1 } }{ { b }_{ 3 } } \)
⇒ \(\frac { x+2 }{ -4 } =\frac { y-3 }{ 5 } =\frac { z-4 }{ -6 } \)
[∵ (x₁, y₁, z₁) is (-2, 3, 4) & (b₁, b₂, b₃) is (-4, -5, -6)]

Given \(\overset { \rightarrow }{ a } .\overset { \rightarrow }{ b } =\overset { \rightarrow }{ a } .\overset { \rightarrow }{ c } =0\)] ⇒ \(\overset { \rightarrow }{ a } \) 丄 \(\overset { \rightarrow }{ b } \)and \(\overset { \rightarrow }{ a } \)丄 \(\overset { \rightarrow }{ c } \) 
⇒ \(\overset { \rightarrow }{ a } \) 丄^r to the plane containing \(\overset { \rightarrow }{ b } \)and \(\overset { \rightarrow }{ c } \) 
Also, \(\overset { \rightarrow }{ b } \times \overset { \rightarrow }{ c } =\left| \overset { \rightarrow }{ b } \right| \left| \overset { \rightarrow }{ c } \right| \) Since \(\overset { \wedge }{ n } \) [where θ is the angle between \(\overset { \rightarrow }{ b } \)and \(\overset { \rightarrow }{ c } \)]
= 1 \(\times\) 1. sin \(\frac { \pi }{ 6 } \).\(\overset { \rightarrow }{ a } \)
[Since \(\overset { \rightarrow }{ b } \) and \(\overset { \rightarrow }{ c } \) are unit vectors \(\overset { \rightarrow }{ a } \) 丄 both \(\overset { \rightarrow }{ b } \)and \(\overset { \rightarrow }{ c } \)  \(\overset { \rightarrow }{ { n } } \) = \(\overset { \rightarrow }{ a } \)]
\(=\frac { 1 }{ 2 } \overset { \rightarrow }{ a } \)
\( \overset { \rightarrow }{ b } \times \overset { \rightarrow }{ c } =\frac { 1 }{ 2 } \overset { \rightarrow }{ a } \)
\(\Rightarrow \overset { \rightarrow }{ a } =\pm 2\left( \overset { \rightarrow }{ b } \times \overset { \rightarrow }{ c } \right) \)

Since the plane passing through the point \(\overset { \rightarrow }{ a } =\overset { \wedge }{ i } -\overset { \wedge }{ j } +2\overset { \wedge }{ k } \) and is normal to the vector \(\overset { \rightarrow }{ n } =2\overset { \wedge }{ i } +3\overset { \wedge }{ j } +2\overset { \wedge }{ k } \)
the vector equation of the plane is \(\overset { \rightarrow }{ r } .\overset { \rightarrow }{ n } =\overset { \rightarrow }{ a } .\overset { \rightarrow }{ n } \)
\(\Rightarrow { \overset { \rightarrow }{ r } }.\left( 2\overset { \wedge }{ i } +3\overset { \wedge }{ j } +2\overset { \wedge }{ k } \right) =\left( \overset { \wedge }{ i } -\overset { \wedge }{ j } +2\overset { \wedge }{ k } \right) .\left( 2\overset { \wedge }{ i } +3\overset { \wedge }{ j } +2\overset { \wedge }{ k } \right) \)
\(\Rightarrow { \overset { \rightarrow }{ r } }.\left( 2\overset { \wedge }{ i } +3\overset { \wedge }{ j } +2\overset { \wedge }{ k } \right) \)= 2 - 3 + 4 = 3
\(\therefore { \overset { \rightarrow }{ r } }.\left( 2\overset { \wedge }{ i } +3\overset { \wedge }{ j } +2\overset { \wedge }{ k } \right) \)= 3

Given (x₁, y₁, z₁) is (2, -1, 3) (x₂, y₂, z₂) is (4, 2, 1)
Cartesian equation of a line passing through two points is \(\frac { x-{ x }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } =\frac { y-{ y }_{ 1 } }{ { y }_{ 2 }-{ y }_{ 1 } } =\frac { z-{ z }_{ 1 } }{ { z }_{ 2 }-{ z }_{ 1 } } \)
\(\Rightarrow \frac { x-2 }{ 2 } =\frac { y+1 }{ 3 } =\frac { z-3 }{ -2 } \)

\(\overset { \rightarrow }{ a } =\overset { \wedge }{ i } +2\overset { \wedge }{ j } +3\overset { \wedge }{ k } \), \(\overset { \rightarrow }{ b } =-\overset { \wedge }{ i } +2\overset { \wedge }{ j } +\overset { \wedge }{ k } \) and \(\overset { \rightarrow }{ c } =3\overset { \wedge }{ i } +\overset { \wedge }{ j } \) 
Since  \(\overset { \rightarrow }{ a } +\lambda \overset { \rightarrow }{ b } \) ⏊^r to \(\overset { \rightarrow }{ c } \)
\(\Rightarrow \left[ \overset { \wedge }{ i } +\overset { \wedge }{ 2j } +\overset { \wedge }{ 3k } +\lambda \left( \overset { \wedge }{ -i } +\overset { \wedge }{ 2j } +\overset { \wedge }{ k } \right) \right] .\left( \overset { \wedge }{ 3i } +\overset { \wedge }{ j } \right) =0\)
\(\Rightarrow \left[ \left( 1-\lambda \right) \overset { \wedge }{ i } +\left( 2+2\lambda \right) \overset { \wedge }{ j } +\left( 3+\lambda \right) \overset { \wedge }{ k } \right] .\left( \overset { \wedge }{ 3i } +\overset { \wedge }{ j } \right) =0\)
⇒ 3(1 - λ) + (2 + 2λ) = 0 ⇒ 3 - 3λ + 2+ 2λ = 0
⇒ 5 - λ = 0 ⇒ λ = 5

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
\(\theta =cos^{ -1 }\left( \frac { |\vec { { n }_{ 1 } } .\vec { { n }_{ 2 } } | }{ |\vec { { n }_{ 1 } } .\vec { { n }_{ 2 } } | } \right) =cos^{ -1 }\left( \frac { |((2\hat { i } +2\hat { j } +2\hat { k } ).(4\hat { i } -2\hat { j } +2\hat { k } ))| }{ |(2\hat { i } +2\hat { j } +2\hat { k } )||4\hat { i } -2\hat { j } +2\hat { k } | } \right) =cos^{ -1 }\left( \frac { \sqrt { 2 } }{ 3 } \right) \).

Length of perpendicular from \(\left( { x }_{ 1 },{ y }_{ 1 },{ z }_{ 1 } \right) \) to the plane.
\(ax+by+cz-p=0\left| \frac { { ax }_{ 1 }+{ by }_{ 1 }+{ cz }_{ 1 }-p }{ \sqrt { { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 } } } \right| \)
\(\therefore \) Length of perpendicular from (1, -2, 3) to the plane
\(x-y-z-5=0\ is\ \delta =\left| \frac { 1-(-2)+3-5 }{ \sqrt { { 1 }^{ 2 }+\left( -1 \right) ^{ 2 }+{ 1 }^{ 2 } } } \right| \)
= \(\left| \frac { 1+2+3-5 }{ \sqrt { 1+1+1 } } \right| =\left| \frac { 1 }{ \sqrt { 3 } } \right| \)
= \(\frac { 1 }{ \sqrt { 3 } } \)