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 \({{x}\over{\alpha}}+{{y}\over{\beta}}+{{z}\over{\gamma}}=3\)

\(\hat { r } .(\hat { i } -2\hat { j } -\hat { k } )\)
\(\overset { \rightarrow }{ a } =2\hat { i } -\hat { j } -\hat { k } \)
\(\overset { \rightarrow }{ n } =\hat { i } -2\hat { j } -4\hat { k } ,d=9\)
Distance = \(\frac { \left| \overset { \rightarrow }{ a. } \quad \overset { \rightarrow }{ n. } \quad -d \right| }{ \overset { \rightarrow }{ n } } \)
\(\frac { \left| (2\hat { i } +\hat { j } -\hat { k } )+4\hat { k } -9 \right| }{ \sqrt { 1+4+16 } } \)
\(\Rightarrow \)  Required distance = \(\frac { \left| 2-2-4-9 \right| }{ \sqrt { 21 } } \)
= \(\frac { 12 }{ \sqrt { 21 } } \) units

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

Intercepts are (3, 0, 0), (0, - 4,0), (0,0,2)
\(\overset { \rightarrow }{ a } =3\check { i } ,b,\overset { \rightarrow }{ b } =-4\overset { \rightarrow }{ j } ,\overset { \rightarrow }{ c } =2\check { k } \)
Vector equation of the plane passing through \(\overset { \rightarrow }{ a } ,\overset { \rightarrow }{ b } \) and \(\overset { \rightarrow }{ c } \) is
\((\overset { \rightarrow }{ r } -\overset { \rightarrow }{ a } ).[(\overset { \rightarrow }{ b } -\overset { \rightarrow }{ a } )\times (\overset { \rightarrow }{ c } -\overset { \rightarrow }{ a } )]=0\)
Here,
\(.(\overset { \rightarrow }{ b } -\overset { \rightarrow }{ a } )\times (\overset { \rightarrow }{ c } -\overset { \rightarrow }{ a } )=(-4\check { j } -3\check { i } )=\times (2\check { k } -3\check { i } )\)
\(=\left| \begin{matrix} \check { i } & \check { j } & \check { k } \\ -3 & -4 & 0 \\ -3 & 0 & 2 \end{matrix} \right| \)
= \(-8\check { i } +6\check { j } -12\check { k } \)

Here \(2\check { i } -3\check { j } +6\check { k } \) , then 
\(|\check { n } |=\sqrt { 4+9+36 } =\sqrt { 49 } \)
=7 units 
\(\check { n } =\frac { \check { n } }{ |\check { n } | } =\frac { 2 }{ 7 } \check { i } -\frac { 3 }{ 7 } \check { j } +\frac { 6 }{ 7 } \check { k } \)
So, the required equation is
\(r\left( \frac { 2 }{ 7 } \check { i } -\frac { 3 }{ 7 } \check { j } +\frac { 6 }{ 7 } \check { k } \right) \)

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

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

\(\overset { \rightarrow }{ r } =(2\overset { \wedge }{ i } +\overset { \wedge }{ j } +3\overset { \wedge }{ k } +\lambda (-2\overset { \wedge }{ i } -3\overset { \wedge }{ j } +4\overset { \wedge }{ k } )\)

Getting 
\(3\overset { \rightarrow }{ a } +2\overset { \rightarrow }{ b } =7\overset { \wedge }{ i } -5\overset { \wedge }{ j } -4\overset { \wedge }{ j } +5\overset { \wedge }{ k } \)
D.R'S are 7, -5, 4

Let points be A(2, 3, 4), B(- 1, - 2, 1) and C(5, 8, 7)
Direction ratios or AB are 2 + 1, 3 + 2, 4 - 1, i.e. 3, 5, 3;
Direction ratios of Be are 5 + 1, 8 + 2, 7 - 1, i.e. 3,5,3
As \(\frac{3}{3}=\frac{5}{5}=\frac{3}{3}\)
⇒  AB is parallel to BC, B is common.
Hence, A, B, C are collinear.

AB

Direction-angles are \(90°,135°,45°\)
Direction-cosines are \(\left< \cos { 90° } ,\cos { 135° } ,\cos { 45° } \right> \)
\(<0,-\frac { 1 }{ \sqrt { 2 } } ,\frac { 1 }{ \sqrt { 2 } } >\)

Let the equation of the plane be
\(\frac { x }{ a } +\frac { y }{ b } +\frac { z }{ c } =1\)
Here a = 2, b = 3, c = 4.
Substituting the values of a, b and c in (1), we get the required equation of the plane as
\(\frac { x }{ 2 } +\frac { y }{ 3 } +\frac { z }{ 4 } =1\Rightarrow 6x+4y+3z=12\)

Let \(\vec { a } \)and \(\vec { b } \) be the position vectors of the points: A(-1,0,2) and B(3,4,6) respectively
\(\text {Then }\vec{a}=-\hat{i}+2 \hat{k} \)
\(\text {and }\vec{b}=3 \hat{i}+4 \hat{j}+6 \hat{k} \)
\(\therefore \vec{b}-\vec{a}=4 \hat{i}+4 \hat{j}+4 \hat{k}\)
Let \(\vec{r}\) be the position vector of any point on the line. Then the vector equation of the line is
\(\vec{r}=-\hat{i}+2 \hat{k}+\lambda(4 \hat{i}+4 \hat{j}+4 \hat{k})\)

Direction cosines are: \(\left< \cos { 90° } ,\cos { 60° } ,\cos { 30° } \right> (i.e.,<0,\frac { 1 }{ 2 } ,\frac { \sqrt { 3 } }{ 2 } >)\)

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

Equation of plane through points A, B and C is 
\(\left| \begin{matrix} x-2 & y-5 & z+3 \\ -4 & -8 & 8 \\ 3 & -2 & 0 \end{matrix} \right| =0\Rightarrow 16x+24y+32z-56=0\)
i.e., 2x + 3y + 4z-7 = 0
Distance of plane from (7, 2, 4)
= \(\left| \frac { 2\left( 7 \right) +3\left( 2 \right) +4\left( 4 \right) -7 }{ \sqrt { 4+9+16 } } \right| =\sqrt { 29 } \)

Let equation of plane through (1, 2, -4) be a(x-1) + b(y-2) + c(z+4) = 0.
The plane is parallel to the given lines
\(\therefore\)  2a + 3b + 6c = 0; a + b - c = 0
Solving: \(\frac { a }{ -9 } =\frac { b }{ 8 } =\frac { c }{ -1 } =k\left( say \right) \)
\(\therefore\) a = -9k, b = 8k, c = -k
From (i), -9k(x-1) + 8k(y-2) - k(z+4) = 0
\(\therefore\) Equation of plane in cartesian form is 9x - 8y + z+11 = 0
Vector form of plane is: \(\Rightarrow \vec { r } .\left( 9\hat { i } -8\hat { j } +\hat { k } \right) =-11\)
Distance of (9, -8, -10) from the plane = \(\left| \frac { 9.9-8\left( -8 \right) +1\left( -10 \right) +11 }{ \sqrt { 81+64+1 } } \right| =\sqrt { 146 } \)