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

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 
\(\Rightarrow \overset { \rightarrow }{ r } \frac { (3\check { i } +2\check { j } +6\check { k } ) }{ 7 } =5\)
\(\Rightarrow \check { r } .(3\check { i } +2\check { j } +6\check { k } )=35\)

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

\(\alpha =60^{ \circ },\beta =?,\gamma =45^{ \circ }\)
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 \(\frac { \pi }{ 3 } \)

\( \alpha=\beta=\gamma \)
\(\Rightarrow 3 \cos ^2 \alpha=1 \)
\(D C^{\prime} s: \pm \frac{1}{\sqrt{3}} \pm \frac{1}{\sqrt{3}} \pm \frac{1}{\sqrt{3}}\)

Point through which line passes is (2, 1, - 4) and dr's: 1, 1,-1.
\(\therefore\) Cartesian equation of line is \(\frac{z-2}{1}=\frac{y-1}{-1}=\frac{z+4}{-1}\)

The given plane is 2x - y + z + 3 = 0 ...(1)
 
Let P (1, 3, 4) be the given point.
Let M be the foot of perpendicular from P on plane (1),
Let P' be the image of P in the plane (1),
The equations of PM are
\(\frac { x-1 }{ 2 } =\frac { y-3 }{ -1 } =\frac { z-4 }{ 1 } \).(2)
Any point on (2) is (1 + 2k, 3 - k, 4 + k) ...(3)
This point is M if it lies on (1)
if 2 (1 + 2k) - (3 - k) + (4 + k) + 3 = 0
if  6k = - 6 if k = - 1.
Putting in (3), the point M is :
(1 + 2 (- 1), 3 - (- 1),4 - 1) i.e. (- 1, 4, 3).
Hence, the co-ordinates of M, the foot of perpendicular
are (- 1, 4, 3). .
And perpendicular distance = I PM I
\(=\sqrt { { \left( 1+1 \right) }^{ 2 }+{ \left( 3-4 \right) }^{ 2 }+{ \left( 4-3 \right) }^{ 2 } } \)
\(=\sqrt { 4+1+1 } =\sqrt { 6 } units\)
Now M is the mid-point of [PP'].
If P' is \((\alpha ,\beta ,\gamma )\), then:
\(\frac { 1+\alpha }{ 2 } =-1,\frac { 3+\beta }{ 2 } =4,\frac { 4+\gamma }{ 2 } =3\)
\(\Rightarrow 1+\alpha =-2,3+\beta =8,4+\gamma =6\)
\(\Rightarrow \alpha =-3,\beta =5,\gamma =2.\)
Hence, the required image is P'(-3, 5, 2)

The x-axis makes angles 0°, 90° and 90° respectively with x, y and z-axis.
Therefore, the direction cosines of x-axis are cos 0°, cos 90°, cos 90° i.e., 1, 0, 0.
Similarly, direction cosines of y-axis and z-axis are 0, 1, 0 and 0, 0, 1 respectively.

We know that the direction-cosines of the line joining P(x₁, y₁, z₁) and q(x₂, y₂, z₂) are:
\(\frac{x_{2}-x_{1}}{\mathrm{PQ}}, \frac{y_{2}-y_{1}}{\mathrm{PQ}}, \frac{z_{2}-z_{1}}{\mathrm{PQ}} \)
\(\text {where } \mathrm{PQ}=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}+\left(z_{2}-z_{1}\right)^{2}} \)
Here P is (– 2, 4, – 5) and Q is (1, 2, 3).
\(\mathrm{PQ}=\sqrt{(1-(-2))^{2}+(2-4)^{2}+(3-(-5))^{2}}=\sqrt{77}\)
Thus, the direction cosines of the line joining two points is
\(\frac{3}{\sqrt{77}}, \frac{-2}{\sqrt{77}}, \frac{8}{\sqrt{77}}\)

General point on the line \(\frac{x+2}{3}=\frac{y+1}{2}=\frac{z-3}{2}=\lambda\)
\( A(3 \lambda-2,2 \lambda-1,2 \lambda+3)\)
Distance of point  A from point (1, 2, 3) is 3 \( \sqrt{2} \text {. }\)
\(\therefore \sqrt{(3 \lambda-2-1)^{2}+(2 \lambda-1-2)^{2}+(2 \lambda+3-3)^{2}} \)
\(=3 \sqrt{2}\)
Squaring and solving for \( \lambda \text {, we get } \lambda=0 \text { or } \lambda=\frac{30}{17}\)
Substituting in (i), we get point as A(-2, -1, 3) or A\(\left( \frac { 56 }{ 17 } ,\frac { 43 }{ 17 } ,\frac { 111 }{ 17 } \right) \) 

Equation of given line is :
\(\overrightarrow{r}=(\hat{i}+\hat{j})+\lambda(\hat{i}+2\hat{j}-\hat{k})=\overrightarrow{a}+\lambda\overrightarrow{b}\) ...(i)
where \(\overrightarrow{a}=\hat{i}+\hat{j}\)
\(\overrightarrow{b}=\hat{i}+2\hat{j}-\hat{k}\)
Again, \(\overrightarrow{r}=(\hat{i}+\hat{j})+\mu(-\hat{i}+\hat{j}-2\hat{k})\)
\(=\vec{a}+\mu\vec{b'}\)
where  \(\overrightarrow{a}=\hat{i}+\hat{j}\)
\(\overrightarrow{b}=-\hat{i}+\hat{j}-2\hat{k}\)
\(\therefore\) The vector equation of the plane containing the line (i) and (ii) is given by 
\(\overrightarrow{b}\times\overrightarrow{b'}=\begin{vmatrix} \hat { i } & \hat { j } & \hat { k } \\ 1 & 2 & -1 \\ -1 & 1 & -2 \end{vmatrix}\)
\(\hat{i}(-4+1)-\hat{j}(-2-1)+\hat{k}(1+2)-3\hat{i}+3\hat{j}+3\hat{k}\)
\(\overrightarrow{n}={{x-8}\over{7}}={{y-4}\over{1}}={{z-5}\over{3}}\)
\(r.\overrightarrow{n}=\overrightarrow{a}.\overrightarrow{n}\)
\(\Rightarrow\)   \(r.(-3\hat{i}+3\hat{j}+3\hat{k})=(\hat{i}+\hat{j})(-3\hat{i}+3\hat{j}+3\hat{k})\)
\(\Rightarrow\)   \(r.(-3\hat{i}+3\hat{j}+3\hat{k})=-3+3=0\)
\(\Rightarrow\)   -3x + 3y + 3z = 0
i.e, x + y + z = 0
\(\bot \) distance from (2, 1, 4) is
\(\left| \frac { 2-1-4 }{ \sqrt { { 1 }^{ 2 }+{ 1 }^{ 2 }+{ 1 }^{ 2 } } } \right| =\frac { 3 }{ \sqrt { 3 } } =\sqrt { 3 } \)

The equation of the planes through the intersection of the plane \(\overrightarrow{r}.(\hat{i}+3\hat{j})-6=0\) and \(\overrightarrow{r}.(3\hat{i}-\hat{j}-4\hat{k})=0\)
\([\overrightarrow{r}.(\hat{i}+3\hat{j})-6]+\lambda[\overrightarrow{r}.(3\hat{i}-\hat{j}-4\hat{k})]=0\)
\(\Rightarrow\)    \(\overrightarrow{r}.[(1+3\lambda)\hat{i}+(3-\lambda)\hat{j}-4\lambda\hat{k}]-6=0\)
This-plane is at a unit distance from the origin.
Therefore, the length of the perpendicular from the origin on (i) is 1unit
\(\Rightarrow\) \({{6}\over{\sqrt{{(1+3\lambda)}^{2}+{(3-\lambda)}^{2}+16{\lambda}^{2}}}}=1\)
\(\Rightarrow\)  \(36={(1+3\lambda)}^{2}+{(3-\lambda)}^{2}+16{\lambda}^{2}\)
\(=1+9{\lambda}^{2}+6\lambda+9+{\lambda}^{2}-6\lambda+16{\lambda}^{2}\)
\(\Rightarrow\)  \(36=26{\lambda}^{2}+10\)
\(\Rightarrow\) \(26{\lambda}^{2}=26\) 
\(\Rightarrow\) \(\lambda=\pm1\)
Put the value of \(\lambda\) in eq(i), then for \(\lambda=1\)
\((4\hat{i}+2\hat{j}-4\hat{j})-6=0\)
and for \(\lambda=-1 \)
\(\overrightarrow{r}.(-2\hat{i}+4\hat{j}+4\hat{k})-6=0\)
These are the equations of the required plane.

Let equation of plane through (1,-1,2) with d'r's of perpendicular as a, b, and c is
a(x -1) + b(y + 1) + c(z - 2) = 0
plane is \(\bot\)to 2x + 3y - 2z = 5
and  x + 2y - 3z = 8
\(\therefore\) 2a + 3b - 2 c= 0
\({{a}\over{-5}}={{b}\over{4}}={{c}\over{1}}=k\)
\(\Rightarrow\)  a = -5k, b = 4k, c = k
Equation of the plane is
- 5k ( x - 1 ) + 4k ( y + 1 ) + k ( z - 2 ) = 0
\(\Rightarrow\)  - 5x + 4y + z + 7 = 0
Distance of plane from (- 2,5,5) is
d = \(\left|{{10+20+5+7}\over{\sqrt{25+16+1}}} \right| \)
\(={{42}\over{\sqrt{42}}}=\sqrt{42}\)

Any point on the line\(\frac { x-1 }{ 2 } =\frac { y+1 }{ 3 } =\frac { z-1 }{ 4 } \) is\(\left( 2\lambda +1,\quad 3\lambda -1,\quad 4\lambda +1 \right) \) .
If the point intersect with the second line, then we have 
\(\frac { 2\lambda +1-3 }{ 1 } =\frac { 3\lambda -1-k }{ 2 } =\frac { 4\lambda +1 }{ 1 } \)
\(\Rightarrow \ \lambda =-\frac { 3 }{ 2 } ,\ hence\ k=\frac { 9 }{ 2 } \)
Equation of plane containing these lines
\(\left| \begin{matrix} x-1 & y+1 & z-1 \\ 2 & 3 & 4 \\ 1 & 2 & 1 \end{matrix} \right| =\)
\(\Rightarrow\) -5(x-1) + 2(y+1)+1(z-1) = 0
i.e., 5x-2y-z-6 = 0