We know that the equation of the plane having intercepts a, band c on the three
co-ordinate axes is
\({{x}\over{a}}+{{y}\over{b}}+{{y}\over{c}}=1\)
Here, the co-ordinates of A, Band C are (a, 0, 0),  (0, b, 0) and (0, 0, c) respectively.
The centroid of \(\triangle \)ABC is \(\left({{a}\over{3}},{{b}\over{3}},{{c}\over{3}}\right).\)
Equating \(\left( {{a}\over{3}},{{b}\over{3}},{{c}\over{3}} \right)\) to \((\alpha,\beta,\gamma),\) we get a = \(3\alpha,\) b = \(3\beta\) and c = \(3\gamma\)
Thus, the equation of the plane is
\({{x}\over{3\alpha}}+{{y}\over{3\beta}}+{{z}\over{3\gamma}}=1\)
or \({{x}\over{\alpha}}+{{y}\over{\beta}}+{{z}\over{\gamma}}=3\)

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
\(\overset { \rightarrow }{ n } =2\check { i } +\check { j } +2\check { k } \)
\(sin\theta =\frac { \overset { \rightarrow }{ b } .\overset { \rightarrow }{ n } }{ \left| \overset { \rightarrow }{ b } \right| \left| \overset { \rightarrow }{ n } \right| } \)
\(=\frac { (6\check { i } +2\check { j } +3\check { k } )(2\check { i } -\check { j } +2k) }{ \sqrt { 39+4+9 } \sqrt { 4+1+4 } } \)
\(=\frac { 12-2+6 }{ \sqrt { 49 } \sqrt { 9 } } \)
\(\Rightarrow sin\theta =\frac { 16 }{ 7\times 3 } =\frac { 16 }{ 21 } \)
\(\therefore \theta =sin^{ -1 }\left( \frac { 16 }{ 21 } \right) \)

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

Normal vector of the plane is
\(\overset { \rightarrow }{ x } =3\check { i } -2\check { j } +6\check { k } \)
\(\left| \overset { \rightarrow }{ x } \right| =\sqrt { 3^{ 2 }+2^{ 2 }+6^{ 2 } } \)
\( =\sqrt { 9+4+36 } =\sqrt { 49 } =7\)
\(\overset { \rightarrow }{ x } =\frac { \overset { \rightarrow }{ x } }{ \left| \overset { \rightarrow }{ x } \right| } =\frac { 3\check { i } -2\check { j } +6\check { k } }{ 7 } \)
The required equation of plane 
\(\Rightarrow \overset { \rightarrow }{ r } \frac { (3\check { i } +2\check { j } +6\check { k } ) }{ 7 } =5\)
\(\Rightarrow \check { r } .(3\check { i } +2\check { j } +6\check { k } )=35\)

Given, equations of line is \(\frac{x+3}{3}=\frac{y-4}{5}=\frac{8-z}{-6}\) or \(\frac{x+3}{3}=\frac{y-4}{5}=\frac{z-8}{6}\) and the point is (-2, 4, -5)
Now, cartesian equations of the required line is \(\frac{x+2}{3}=\frac{y-4}{5}=\frac{z+5}{6}\)   ...(i)
[\(\because\) in parallel lines, direction ratios are proportional] and vector equation of the required line is
\(\vec{r}=(-\hat{2} i+\hat{4} j-\hat{5} k)+\lambda(3 \hat{i}+5 \hat{j}+6 \hat{k})\)

Position vector of the point (1, 2, 3)
\(\therefore \overset { \rightarrow }{ a } =\overset { \wedge }{ i } +2\overset { \wedge }{ j } +3\overset { \rightarrow }{ k } \)
and parallel to vector \(3\overset { \wedge }{ i } +2\overset { \wedge }{ j } -3\overset { \wedge }{ k } \)
Required vector equation of line
\(\overset { \rightarrow }{ r } =\overset { \rightarrow }{ a\quad } +\lambda \overset { \rightarrow }{ b\quad } \)
\(\Rightarrow \overset { \rightarrow }{ r } =\overset { \rightarrow }{ i } +2\overset { \rightarrow }{ j } +3\overset { \rightarrow }{ k } +l(3\overset { \rightarrow }{ i } +2\overset { \rightarrow }{ j } -3\overset { \rightarrow }{ k } )\)
Cartesian equation
\(\frac { x-1 }{ 3 } =\frac { y-2 }{ 2 } =\frac { z-2 }{ -3 } \)

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
\(a_{ 1 },a_{ 2 }+b_{ 1 },b_{ 2 },+c_{ 1 },c_{ 2 }=0\)
\(\Rightarrow 3\times 2+2\times 5+4\times -4=0\)
\(\Rightarrow 6+10-16=0\)
\(\Rightarrow 16-16=0\)
\(\therefore AB\bot to\quad CD\)

Since ABCD is a parallelogram and diagonal are bisect each other.
Let A(3, - 1, - 1), B(5, - 4, 0), C(2, 3, - 2), \(D(x_{ 2 },y_{ 2 },z_{ 2 })\)
Midpoint of AC \(\left( \frac { 2+3 }{ 2 } ,\frac { 3-1 }{ 2 } .\frac { -2-1 }{ 2 } \right) \)
\(=\left( \frac { 5 }{ 2 } ,\frac { 2 }{ 2 } ,\frac { -3 }{ 2 } \right) \)
\(=\left( \frac { 5 }{ 2 } ,1,\frac { -3 }{ 2 } \right) \)
Mid-point of BD is same let P be the mid-point of ACand BD
\(\frac { 5 }{ 2 } =\frac { 5+x_{ 2 } }{ 2 } ,1=\frac { -4+y_{ 2 } }{ 2 } \)
\(\frac { -3 }{ 2 } =\frac { 0+z_{ 2 } }{ 2 } \)
\(\Rightarrow x_{ 2 }=0,y_{ 2 }=6,z_{ 2 }=-3\)
D(0,6,-3)

\(\overset { \rightarrow }{ b } _{ 1 }=-i+j+3k\)
and \(\overset { \rightarrow }{ b } _{ 2 }=(2i-3j+k)\)
\(cos\theta =\frac { \overset { \rightarrow }{ b_{ 1 }. } \overset { \rightarrow }{ b_{ 2 }. } }{ |\overset { \rightarrow }{ b_{ 1 }. } ||\overset { \rightarrow }{ b_{ 2 }. } | } \)
\(=\frac { (i+j+3k)(2\overset { \wedge }{ i } -3\overset { \wedge }{ j } +k) }{ \sqrt { 1^{ 2 }+1^{ 2 }+3^{ 2 }\sqrt { 2^{ 2 }+3^{ 2 }+1^{ 2 } } } } \)
\(=\frac { (2-3+3) }{ \sqrt { 1+1+9 } \sqrt { 4+9+1 } } \)
\(\Rightarrow cos\theta =\frac { 2 }{ \sqrt { 11 } \sqrt { 14 } } =\frac { 2 }{ \sqrt { 154 } } \)
\(\Rightarrow \theta =cos^{ -1 }\left( \frac { 2 }{ \sqrt { 154 } } \right) \)

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

Given equation
x = ay + b, Z = cy + d
\(\frac { x-b }{ a } =y,\frac { x-d }{ c } =y\)
\(\Rightarrow \frac { x-b }{ z } =\frac { y-0 }{ 1 } =\frac { z-d }{ c } \)
Direction ratios are (a, 1, c) and a point on the given line is (b, 0, d).

\(\frac { x-2 }{ 2 } =\frac { 2y-5 }{ -3 } ,z=-1\)
\(\Rightarrow \frac { x-2 }{ 2 } =\frac { 2y-5 }{ -3 } =\lambda ,z=-1+0\lambda \)
\(\frac { x-2 }{ 2 } =\frac { 2y-5 }{ -3 } ,z=-1\)
\(\Rightarrow \frac { x-2 }{ 2 } =\frac { y-5/2 }{ -3/2 } =\lambda ,z=-1+0\lambda \)
\(\Rightarrow \frac { x-2 }{ 2 } =\frac { y-5/2 }{ -3/2 } =\lambda ,\frac { z+1 }{ 0 } =\lambda \)
ratios are (2, - 3/2, 0) and the point on the given line is(2, 5/2, -1).

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

\(I=cos\alpha ,m=cos\beta ,n=cos\gamma \)
We know that
\(i^{ 2 }+m^{ 2 }+n^{ 2 }=1\)
\(\Rightarrow cos^{ 2 }\alpha +cos^{ 2 }\beta +cos^{ 2 }\gamma =1\)
as we know that ,cos 2x=2 cos² x-1
\(\Rightarrow \left( \frac { 1+cos2\alpha }{ 2 } \right) +\left( \frac { 1+cos2\beta }{ 2 } \right) \left( \frac { 1+cos2\gamma }{ 2 } \right) =1\)
\(\Rightarrow 1+cos2\alpha +1+cos2\beta +1+cos2\gamma =2\)
\(\Rightarrow cos\quad 2\alpha +cos2\beta +1+cos2\gamma =2\)
\(cos\quad 2\alpha +cos2\beta +cos\quad 2\gamma =-1\)

\(\alpha =60^{ \circ },\beta =?,\gamma =45^{ \circ }\)
Let \(\alpha \) makes with x-axis, y makes with z-axis and \(\beta \) makes with y-axis
\(\alpha =cos\quad 60^{ \circ },m=cos\quad \beta ,y=cos45^{ \circ }\)
\(\alpha ^{ 2 }+\beta ^{ 2 }+\gamma ^{ 2 }=1\)
\(\Rightarrow cos^{ 2 }60^{ \circ }+cos^{ 2 }\beta +cos^{ 2 }45^{ \circ }\)
\(\Rightarrow \frac { 1 }{ 4 } +cos^{ 2 }\beta +\frac { 1 }{ 2 } =1\)
\(\Rightarrow cos^{ 2 }\beta =1-\frac { 1 }{ 4 } -\frac { 1 }{ 2 } \)
\(=\frac { 4-1-2 }{ 4 } \)
\(=\frac { 1 }{ 4 } \)
\(\Rightarrow cos \beta =\pm \frac { 1 }{ 2 } \)
\(\beta =\frac { \pi }{ 3 } or\ \frac { 2\pi }{ 3 } \)
But angle between two lines in the interval \(\left( 0,\frac { \pi }{ 2 } \right) \)
Hence required angles is \(\frac { \pi }{ 3 } \)