The coordinates of the points, O and P, are (0, 0, 0) and (1, 2, −3) respectively.
Therefore, the direction ratios of OP are (1 − 0) = 1, (2 − 0) = 2, and (−3 − 0) = −3
It is known that the equation of the plane passing through the point (x₁, y₁ z₁) is 
\(a\left(x-x_{1}\right)+b\left(y-y_{1}\right)+c\left(z-z_{1}\right)=0\) where, a, b, and c are the direction ratios of normal.
Here, the direction ratios of normal are 1, 2, and −3 and the point P is (1, 2, −3).
Thus, the equation of the required plane is
1(z-1) +2(y-2) - 3(z+3) = 0
\(\Rightarrow x+2 y-3 z-14=0\)

The given cartesian equation of the line is:
\(\frac { x-5 }{ 3 } =\frac { y+4 }{ 7 } =\frac { z-6 }{ 2 } \) . (1) 
This passes through the point(5,-4,6) and has <3,7,2> as its direction ratios
\(\Rightarrow\) the line (1) passes through the point A\((\vec { a } )\)where
\(\vec { a } =5\vec { i } -4\vec { j } +6\vec { k } \ \text {and is in the direction of}\)
\( \vec { m } =3\vec { i } +7\vec { j } +2\vec { k } \)
The vector equation of the required line is:
\(\vec { r } =\vec { a } +\lambda \vec { m } \)
\(i.e.\ \vec { r } =\left( 5\vec { i } -4\vec { j } +6\vec { k } \right) +\lambda \left( \vec { i } +7\vec { j } +2\vec { k } \right) \)

It is given that the line passes through the point A (1, 2, 3). Therefore, the position vector through A 
\(\vec { a } =\hat { i } +2\hat { j } +3\hat { k } \quad and\)
\(\vec { b } =3\vec { i } +2\vec { j } -2\vec { k } \)
The equation of the required line is:
\(\vec { r } =\left( \hat { i } +2\hat { j } +3\hat { k } \right) +\lambda \left( 3\vec { i } +2\vec { j } -2\vec { k } \right) ;\lambda \in R\)

\(\text { Here, } \vec{a}=2 \hat{i}+5 \hat{j}-3 \hat{k}, \overrightarrow{\mathrm{N}}=6 \hat{i}-3 \hat{j}+2 \hat{k} \text { and } d=4\)
Therefore, the distance of the point (2, 5, – 3) from the given plane is
\(\frac{|(2 \hat{i}+5 \hat{j}-3 \hat{k}) \cdot(6 \hat{i}-3 \hat{j}+2 \hat{k})-4|}{|6 \hat{i}-3 \hat{j}+2 \hat{k}|}=\frac{|12-15-6-4|}{\sqrt{36+9+4}}=\frac{13}{7}\)

Comparing the given equations of the planes with the equations
\(\mathrm{A}_{1} x+\mathrm{B}_{1} y+\mathrm{C}_{1} z+\mathrm{D}_{1}=0 \text { and } \mathrm{A}_{2} x+\mathrm{B}_{2} y+\mathrm{C}_{2} z+\mathrm{D}_{2}=0\)
\(\text {We get }\mathrm{A}_{1}=3, \mathrm{~B}_{1}=-6, \mathrm{C}_{1}=2\)
\(\mathrm{A}_{2}=2, \mathrm{~B}_{2}=2, \mathrm{C}_{2}=-2 \)
\(\cos \theta =\left|\frac{3 \times 2+(-6)(2)+(2)(-2)}{\sqrt{\left(3^{2}+(-6)^{2}+(-2)^{2}\right) \sqrt{\left(2^{2}+2^{2}+(-2)^{2}\right)}}}\right| \)
\(=\left|\frac{-10}{7 \times 2 \sqrt{3}}\right|=\frac{5}{7 \sqrt{3}}=\frac{5 \sqrt{3}}{21} \)
\(\text {Therefore, }\theta=\cos ^{-1}\left(\frac{5 \sqrt{3}}{21}\right) \)

As line AB is perpendicular to the plane therefore, direction ratios of the line are 2, 3,4 and line passes through the point A( 1, -2, 3)
\(\therefore \) equation of the perpendicular is
\(\frac{x-1}{2}=\frac{y+2}{-3}=\frac{z-3}{4}=\lambda \text { (say) }\)
General point on the line is
\(B(2 \lambda+1,-3 \lambda-2,4 \lambda+3)\)
If the point lies on the plane, then 
\(2(2\lambda +1)-3(-3\lambda -2)+4(4\lambda +3)+9=0\Rightarrow 29\lambda =-29\Rightarrow \lambda =-1\)
Substituting in (i), we get foot of the perpendicular as B(-1, 1, -1).

Plane passing through the point A(0, 0, 0) is
a(x - 0) + bev - 0) + c(z - 0) = 0 ...(i)
Plane (i) passes through the point (3, -1, 2)
3a - b + 2c = 0
Plane (i) is parallel to the line \( \frac{\dot{x}-4}{1}=\frac{y+3}{-4}=\frac{z+1}{7}\)
\(\therefore \quad a-4 b+7 c=0\)
Eliminating a, b, c  from (i), (ii) and (iii), we get
\(\left|\begin{array}{lll} x & y & z \\ 3 & -1 & 2 \\ 1 & -4 & 7 \end{array}\right|=0 \)
\(\Rightarrow x(-7+8)-y(21-2)+z(-12+1)=0 \)
\(\Rightarrow x-19 y-11 z=0 \) is the required equation.

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

\(d_{ 1 }=10,d_{ 2 }=9\)
\(d_{ 0 }=2,b=3,c=4\)
The equation 4x + 6y + 8z = 18 can be written as 
2x + 3y + 4z = 9
Now distance between planes
\(\frac { \left| d_{ 1 }-d_{ 2 } \right| }{ \sqrt { a^{ 2 }+b^{ 2 }+c^{ 2 } } } \)
\(=\frac { \left| 10-9 \right| }{ \sqrt { 4+9+16 } } =\frac { 1 }{ \sqrt { 29 } } \)

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

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
\(\vec{r}=(-\hat{2} i+\hat{4} j-\hat{5} k)+\lambda(3 \hat{i}+5 \hat{j}+6 \hat{k})\)

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)

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 = \(\left| \frac { 9.9-8\left( -8 \right) +1\left( -10 \right) +11 }{ \sqrt { 81+64+1 } } \right| =\sqrt { 146 } \)