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.

Lines \({{x-{x}_{1}}\over{{l}_{1}}}={{y-{y}_{1}}\over{{m}_{1}}}={{z-{z}_{1}}\over{{n}_{1}}}\) and \({{x-{x}_{2}}\over{{l}_{2}}}={{y-{y}_{2}}\over{{m}_{2}}}={{z-{z}_{2}}\over{{n}_{2}}}\) are coplaner, if
\(\begin{vmatrix} { x }_{ 2 }-{ x }_{ 1 } & { y }_{ 2 }-{ y }_{ 1 } & { z }_{ 2 }-{ z }_{ 1 } \\ { l }_{ 1 } & { m }_{ 1 } & { n }_{ 1 } \\ { l }_{ 2 } & { m }_{ 2 } & { n }_{ 2 } \end{vmatrix}=0\)
Now, \(\begin{vmatrix} -1+3 & 2-1 & 5-5 \\ -3 & 1 & 5 \\ -1 & 2 & 5 \end{vmatrix}=\begin{vmatrix} 2 & 1 & 0 \\ -3 & 1 & 5 \\ -1 & 2 & 5 \end{vmatrix}\)
= 2 (5 -10) - 1 (-15 + 5) + 0
= -10 + 10 = 0
Thus, given lines are coplanar.
Now, equation of plane containing these lines is:
\(\begin{vmatrix} x+3 & y-1 & x-5 \\ -3 & 1 & 5 \\ -1 & 2 & 5 \end{vmatrix}=0\)
\(\Rightarrow\) (x + 3) (5 -10) - (y -1) (-15 + 5) + (z - 5) (- 6 + 1) = 0
\(\Rightarrow\) - 5 (x + 3) + 10(y-l) -5(z -5) = 0
\(\Rightarrow\) - 5x + 10y - 5z = 0
\(\Rightarrow\)  x - 2y + z = 0.

Equation of plane: 2x + y + 3z = 26
Normal to the plane = \(2\hat{i}+\hat{j}+3\hat{k}\)
\(\therefore \) < 2, 1, 3 > are direction ratios of the normal to the plane.

Equation of the line through P(2, 3, 4) and perpendicular to the give plane is
\({{x-2}\over{2}}={{y-3}\over{1}}={{z-4}\over{3}}=\lambda\)
\(\therefore\) Co-ordinates of point N are N \((2\lambda+2,\lambda+3,3\lambda+4)\)
Since N lies on the plane,
\(\Rightarrow\)  \(2(2\lambda+2)+(\lambda+3)+3(3\lambda+4)=26\)
\(\Rightarrow\)  \(4\lambda+4+\lambda+3+9\lambda+12=26\)
\(\Rightarrow\)  \(14\lambda=26-19\)
\(\Rightarrow\)  \(14\lambda=7\)
\(\Rightarrow\)  \(\lambda={{1}\over{2}}\)
\(\therefore\) Co-ordinates of the foot of the perpendicular i.e., N are
\(\left( 2\left({{1}\over{2}} \right)+2,{{1}\over{2}}+3,3\left({{1}\over{2}} \right)+4 \right)\) i.e., \(\left( 3,{{7}\over{2}},{{11}\over{2}} \right)\)
\(\therefore\) The length of perpendicular form P to given plane is
\(|NP|=\sqrt{{2-3}^{2}+{\left(3-{{7}\over{2}} \right)}^{2}+\left(4-{{11}\over{2}}\right)^{2}}\)
\(=\sqrt{1+{{1}\over{4}}+{{9}\over{4}}}\)
\(=\sqrt{{{14}\over{1}}}\)
\(=\sqrt{{7}\over{2}}\) units
Now, N is the mid-point of PP', where P'\((\alpha,\beta,\gamma)\) is the image of point P.
\(\therefore\)   \(\left(3,{{7}\over{2}},{{11}\over{2}} \right)=\left({{2+\alpha}\over{2}} ,{{3+\beta}\over{2}},{{4+\gamma}\over{2}}\right)\)
\(\Rightarrow \) \(3={{2+\alpha}\over{2}};{{7}\over{4}}={{3+\beta}\over{2}};{{11}\over{2}}={{4+\gamma}\over{2}}\)
\(\Rightarrow\)  \(6=2+\alpha;7=3+\beta;11=4+\gamma\)
\(\Rightarrow\)  \(\alpha=4;\beta=4;\gamma=7\)
\(\therefore\) Image of point P is P'(4, 4, 7).

Any point on the given line is \(\left( 2\lambda -1,\ -2\lambda +3,\ -\lambda \right) \). In this point is Q then
\(\vec { PQ } =\left( 2\lambda -2 \right) \hat { i } +\left( -2\lambda +1 \right) \hat { j } +\left( -\lambda +3 \right) \hat { k } \)
Since \(\vec { PQ }\) is perpendicular to the line 
\(\therefore 2\left( 2\lambda -2 \right) -2\left( -2\lambda +1 \right) -1\left( -\lambda +3 \right) =0\)
\(\Rightarrow \lambda =1\)
 \(\therefore\)  Foot of perpendicular is Q(1, 1, -1)
Let P'(x, y, z) be the image of P in the line, then
\(\frac { x+1 }{ 2 } =1\),
\(\frac { y+2 }{ 2 } =1\),
\(\frac { y-3 }{ 2 } \) =-1
\(\Rightarrow\) x = 1, y = 0, z = 1
\(\Rightarrow\) Image is (1, 0, 1)

Given, 2x = y = z
\(\Rightarrow\)  \(\frac{x}{1/2}\)=\(\frac{y}{1}\)=\(\frac{z}{1}\)
Direction ratios of this line are \(\left( \frac { 1 }{ 2 } ,1,1 \right) \)
\(\therefore\) Line parallel to this line and passing through (3, 4, 5) is \(\frac{x-3}{1/2}\)=\(\frac{y-4}{1}\)=\(\frac{z-5}{1}\)= k (let) 
\(\Rightarrow\) x = 3 + \(\frac{k}{2}\) ,y = 4 + k , z = 5 + k
Substituting in the plane x + y + z = 2, we get 3 + \(\frac{k}{2}\) + 4 + k + 5 + k = 2, 
\(\Rightarrow\) 12 +\(\frac{5k}{2}\) = 2
\(\Rightarrow\) 5k = -10\(\times\)2​​​​​​​
\(\Rightarrow\) k = -4
Point of intersection is:
\(\left( 3-\frac { 4 }{ 2 } ,\ 4-4,\ 5-4 \right) =\left( 1,\ 0,\ 1 \right) \)
\(\therefore\) Distance = \(\sqrt { { \left( 3-1 \right) }^{ 2 }+{ \left( 4-0 \right) }^{ 2 }+{ \left( 5-1 \right) }^{ 2 } } =\sqrt { 4+16+16 } \)
 = 6 units

Any point on the line \(\vec { r } =\left( 2\hat { i } -4\hat { j } +2\hat { k } \right) +\lambda \left( 3\hat { i } +4\hat { j } +2\hat { k } \right) \)is
\(\left( 2+3\lambda \right) \hat { i } +\left( -4+4\lambda \right) \hat { j } +\left( 2+2\lambda \right) \hat { k } \). 
For the line to intersect the plane, the above point must satisfy the equation of plane, for some value of \(\lambda \).
\(\therefore \left\{ \left( 2+3\lambda \right) \hat { i } +\left( -4+4\lambda \right) \hat { j } +\left( 3+2\lambda \right) \hat { k } \right\} .\left( \hat { i } -2\hat { j } +\hat { k } \right) =0\)
\(\Rightarrow 2+3\lambda +8-8\lambda +2+2\lambda =0\Rightarrow \lambda =4\)
\(\therefore\) The point of intersection is \(14\hat { i } +12\hat { j } +10\hat { k } \).

Required distance = \(\sqrt { { 12 }^{ 2 }+{ 0 }^{ 2 }+{ 5 }^{ 2 } } \) = 13 units.

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)
= \(\left| \frac { 2\left( 7 \right) +3\left( 2 \right) +4\left( 4 \right) -7 }{ \sqrt { 4+9+16 } } \right| =\sqrt { 29 } \)

Any plane through the line of intersection of the given planes is 
\(\vec { r } .\left( 2\hat { i } +3\hat { j } -\hat { k } \right) +1+\lambda \left\{ \vec { r }\left( \hat { i } +\hat { j } -2\hat { k } \right) \right\} =0\)
or \(\vec { r } .\left\{ \left( 2+\lambda \right) \hat { i } +\left( 3+\lambda \right) \hat { j } +\left( -1-2\lambda \right) \hat { k } \right\} =-1\)
If it contains the point (3, -2, -1), then we have(3) (2+\(\lambda\)) + (-2)(3+\(\lambda\)) + (-1)(-1-2\(\lambda\)) = -1.
\(\Rightarrow \lambda =-\frac { 2 }{ 3 } \)
The required equation of the plane is 
\(\vec { r } \left\{ \left( 2-\frac { 2 }{ 3 } \right) \hat { i } +\left( 3-\frac { 2 }{ 3 } \right) \hat { j } +\left( -1+\frac { 4 }{ 3 } \right) \hat { k } \right\} =-1\)
or \(\vec { r } .\left( 4\hat { i } +7\hat { j } +\hat { k } \right) =-3\)
If \(\theta\) be the angle between the normals to the two given planes, then \(\theta\) is the angle between the planes and
\(cos\quad \theta =\frac { \vec { n } _{ 1 }.\vec { n } _{ 2 } }{ \left| \vec { n } _{ 1 } \right| \left| \vec { n } _{ 2 } \right| } \)
\(=\frac { \left( 2\hat { i } +3\hat { j } -\hat { k } \right) \left( \hat { i } +\hat { j } -2\hat { k } \right) }{ \left| \left( 2\hat { i } +3\hat { j } -\hat { k } \right) \right| \left| \left( \hat { i } +\hat { j } -2\hat { k } \right) \right| } \)
\(=\frac { 2+3+2 }{ \left( \sqrt { { 2 }^{ 2 }+{ 3 }^{ 2 }+{ 1 }^{ 2 } } \right) \left( \sqrt { { 1 }^{ 2 }+{ 1 }^{ 2 }+{ 2 }^{ 2 } } \right) } =\frac { 7 }{ 2\sqrt { 21 } } \)

The equation of the line passing through point (3, -2, 1) and parallel to the given line is \(\frac { x-3 }{ 2 } =\frac { y+2 }{ -3 } =\frac { z-1 }{ 1 } =\lambda\)
Any point on this line, we have (2\(\lambda \)+3, -3\(\lambda \)-2, \(\lambda \)+1)
If it lies on the plane, we have 3(2\(\lambda \)+3)-3\(\lambda \)-2-\(\lambda \)-1+2 = 0
\(\Rightarrow \lambda =-4\)
Hence, the point common to the plane and the line is (-5, 10, -3).
Hence the required distance = \(\sqrt { { \left( 3+5 \right) }^{ 2 }+{ \left( -2-10 \right) }^{ 2 }+{ { \left( 1+3 \right) } }^{ 2 } } =4\sqrt { 14 } units\)
The equation of the line passing through (3, -2, 1) and perpendicular to the plane is \(\frac { x-3 }{ 3 } =\frac { y+2 }{ 1 } =\frac { z-1 }{ -1 } =\mu\)
Any point on it is (3\(\mu\)+3, \(\mu\)-2, -\(\mu\) +1)
If it lies on the plane, we get 3(3\(\mu\)+3)+\(\mu\) -2+ \(\mu\)-1+2 = 0
\(\Rightarrow\) \(\mu=\frac{8}{11}\)
The required foot of the perpendicular is \(\left( \frac { 9 }{ 11 } ,\ \frac { -30 }{ 11 } ,\ \frac { 19 }{ 11 } \right) \).

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