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'
\(cos\theta =\frac { { \vec { n } }_{ 1 }.{ \vec { n } }_{ 2 } }{ \left| { \vec { n } }_{ 1 } \right| \left| { { \vec { n } }_{ 2 } } \right| } =\frac { \left| 1(2)+1(-2)-2(1) \right| }{ \sqrt { { 1 }^{ 2 }+{ 1 }^{ 2 }+\left( -2 \right) ^{ 2 }.\sqrt { { 2 }^{ 2 }+\left( -2 \right) ^{ 2 }+{ 1 }^{ 2 } } } } \)
= \(\frac { \left| -2 \right| }{ \sqrt { 6 } .\sqrt { 9 } } =\frac { 2 }{ \sqrt { 6 } (3) } =\frac { 2 }{ 3\sqrt { 6 } } \)
\(\theta ={ { cos }^{ -1 }\left( \frac { 2 }{ 3\sqrt { 6 } } \right) }\)

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
\(x-y-z-5=0\ is\ \delta =\left| \frac { 1-(-2)+3-5 }{ \sqrt { { 1 }^{ 2 }+\left( -1 \right) ^{ 2 }+{ 1 }^{ 2 } } } \right| \)
= \(\left| \frac { 1+2+3-5 }{ \sqrt { 1+1+1 } } \right| =\left| \frac { 1 }{ \sqrt { 3 } } \right| \)
= \(\frac { 1 }{ \sqrt { 3 } } \)