(i) The given lines are:
 \(\vec { r } =3\hat { i } +2\hat { j } -4\hat { k } +\lambda (\hat { i } +2\hat { j } +2\hat { k } )\)
\(\vec { r } =5\hat { i } -2\hat { j } +\mu (3\hat { i } +2\hat { j } +6\hat { k } )\)
 \(i.e.\ \vec { r } =(3+\lambda )\hat { i } +(2+2\lambda )\hat { j } +(-4+2\lambda )\hat { k } ....(1)\)
\(and\quad \vec { r } =(5+3\mu )\hat { i } +(-2+2\mu )\hat { j } +6\mu \hat { k }..(2)\)
If the lines (1) and (2) intersect, then for some values of \(\lambda \ and\ \mu \)we have :
\(3+\lambda =5+3\mu ..(3)\)
\(2+2\lambda =-2+2\mu ...(4)\)
\(-4+2\lambda =6\mu ...(5)\)
Solving (4) and (5),\(\lambda =-4\ and\ \mu =-2\)
These satisfy(1). [3-4 = 5-6 i.e.-1 = -1]
Hence the lines intersect.
Putting \(\lambda =-4\)in (1)
\(\vec { r } =(3-4)\hat { i } +(2-8)\hat { j } +(-4-8)\hat { k } \)
\(=-\hat { i } -6\hat { j } -12\hat { k } \)
Putting \(\mu =-2\)in (2)
 \(\vec { r } =(5-6)\hat { i } +(-2-4)\hat { j } +6(-2)\hat { k } \)
\( =-\hat { i } -6\hat { j } -12\hat { k } \)
Hence, the point of intersection is(-1, -6, -12)

 line is  \(\vec { r } .(\hat { i } -2\hat { j } +3\hat { k } )+\lambda (-3\hat { i } +5\hat { j } +4\hat { k } )\)
In Cartesian from : \(\frac { x-1 }{ -3 } =\frac { y-2 }{ 5 } =\frac { z-3 }{ 4 } \)

Let line through point (2, - 1, 3) is
\(\vec{r}=(2 \hat{i}-\hat{j}+3 \hat{k})+\lambda^{\prime}(a \hat{i}+b \hat{j}+c \hat{k})\)
\(\text { If line }(i) \text { is perpendicular to lines }\)
\(\vec{r}=(\hat{i}+\hat{j}-\hat{k})+\lambda(2 \hat{i}-2 \hat{j}+\hat{k}) \)
\(\text {and } \vec{r}=(2 \hat{i}-\hat{j}-3 \hat{k})+\mu(\hat{i}+2 \hat{j}+2 \hat{k}) \)
\(\text { then }(a \hat{i}+b \hat{j}+c \hat{k}) \cdot(2 \hat{i}-2 \hat{j}+\hat{k})=0\)
\(\Rightarrow 2 a-2 b+c=0\)
\(\text { and }(a \hat{i}+b \hat{j}+c \hat{k}) \cdot(\hat{i}+2 \hat{j}+2 \hat{k})=0 \Rightarrow a+2 b+2 c=0\)
\(\Rightarrow \frac{a}{4-2}=\frac{-b}{4-1}=\frac{c}{4+2}, \text { i.e. } \frac{a}{6}=\frac{b}{-3}=\frac{c}{6} \)
\(\Rightarrow a: b: c \text { is }-6:-3: 6 \text { or } 2: 1:-2 \)
 line is \(\vec { r } =(2\hat { i } +\hat { j } -3\hat { k } )+\lambda '(2\hat { i } -\hat { j } +2\hat { k } )\)

Consider, the line \( \vec{r}=(1+\lambda) \hat{i}+(2-\lambda) \hat{j}+(\lambda+1) \hat{k} \)
\(i.e., \vec{r}=(\hat{i}+2 \hat{j}+\hat{k})+\lambda(\hat{i}-\hat{j}+\hat{k}) \)
\(\text {Here } \vec{a}_{1}=\hat{i}+2 \hat{j}+\hat{k}, \vec{b}_{1}=\hat{i}-\hat{j}+\hat{k} \)
Shortest distance between the lines 
\(=\left|\frac{\left(\overrightarrow{a_{2}}-\vec{a}_{1}\right) \cdot\left(\vec{b}_{1} \times \overrightarrow{b_{2}}\right)}{\left|\vec{b}_{1} \times \overrightarrow{b_{2}}\right|}\right|\)
\(\text {Now } \vec{a}_{2}-\vec{a}_{1}=2 \hat{i}-\hat{j}-\hat{k}-\hat{i}-2 \hat{j}-\hat{k}=\hat{i}-3 \hat{j}-2 \hat{k}\)
\(\vec{b}_{1} \times \vec{b}_{2} =\left|\begin{array}{rrr} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 1 \\ 2 & 1 & 2 \end{array}\right| \)
\(-\hat{i}(-2-1)-\hat{j}(2-2)+\hat{k}(1+2) \)
\(=-3 \hat{i}+3 \hat{k} \)
\(\left|\vec{b}_{1} \times \vec{b}_{2}\right| =\sqrt{9+9}=3 \sqrt{2} \)
Shortest distance = \(\left| \frac { (\acute { i } -3\acute { j } -2\acute { k } ).(-3\acute { i } +3\acute { k } ) }{ 3\sqrt { 2 } } \right| =\left| \frac { -3-6 }{ 3\sqrt { 2 } } \right| =\frac { 3 }{ \sqrt { 2 } } =\frac { 3\sqrt { 2 } }{ 2 } \) units

Direction ratios of normal to the plane are 7, -8, 1.
Substituting in (i), we get 
7(x+1)-8(y-3)+3(z-2) = 0 
\(\Rightarrow \) 7x-8y+3z+25 = 0
which is the required equation.

Plane is \( \vec{r} \cdot\{(\hat{i}+\hat{j}+\hat{k})+\lambda(2 \hat{i}+3 \hat{j}+4 \hat{k})\}=6-5 \lambda\)
Plane passes through point with position vector \( (\hat{i}+\hat{j}+\hat{k}) \)
\(\Rightarrow(\hat{i}+\hat{j}+\hat{k}) \cdot\{(\hat{i}+\hat{j}+\hat{k})+\lambda(2 \hat{i}+3 \hat{j}+4 \hat{k})\}=6-5 \lambda \)
\(\Rightarrow(1+1+1)+\lambda(2+3+4)=6-5 \lambda \)
\(\Rightarrow 3+9 \lambda=6-5 \lambda \Rightarrow 14 \lambda=3 \)
\(\Rightarrow \lambda=\frac{3}{14} \)
Substituting in (i), we get 
Plane as \( \vec{r} \cdot\left\{(\hat{i}+\hat{j}+\hat{k})+\frac{3}{14}(2 \hat{i}+3 \hat{j}+4 \hat{k})\right\}\)
\(=6-5 \times \frac{3}{14}\)
\(\Rightarrow \vec{r} \cdot\{14 \hat{i}+14 \hat{j}+14 \hat{k}+6 \hat{i}+9 \hat{j}+12 \hat{k}\}=84-15 \)
\(\Rightarrow \vec{r} \cdot(20 \hat{i}+23 \hat{j}+26 \hat{k})=69 \)

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.

Required equation of the plane is
\(\vec { r } .[(1-2\lambda )]\hat { i } +(-2+\lambda )\hat { j } +(3+\lambda )\hat { k } ]=4-5\lambda \)
Intercept of the plane on x-axis = Intercept of the plane on y-axis.
\(\frac { 4-5\lambda }{ 1-2\lambda } =\frac { 4-5\lambda }{ 1-2 } \)
i.e  \(\lambda =1,\frac { 4 }{ 5 } (rejecting\quad \lambda =\frac { 4 }{ 5 } )\)
Required equation of the plane is
\(\vec { r } .(-\hat { i } -\hat { j } +4\hat { k } )+1=0\)

Equation of line through (3,- 4,- 5) and (2,- 3,1)
\({ { x-3}\over{2-3 } }={ { y-(-4)}\over{-3-(-4) } }={ { z-(-5)}\over{ 1-(-5)} }\)
\(\Rightarrow\) \({ {x-3 }\over{ -1} }={ {y+4 }\over{1 } }={ { z+5}\over{ 6} }=\lambda\)(Let)
\(\therefore\) Co-ordinate of any random point on this line
M \((-\lambda+3,\lambda-4,6\lambda-5).\)
Also, equation of plane passing through the points (2, 2,1), (3, 0,1) and (4, -1, 0) is given by the formula:
\(\begin{vmatrix} x-{ x }_{ 1 } & y-{ y }_{ 1 } & z-{ z }_{ 1 } \\ { x }_{ 2 }-{ x }_{ 1 } & { y }_{ 2 }-{ y }_{ 1 } & { z }_{ 2 }-{ z }_{ 1 } \\ { x }_{ 3 }-{ x }_{ 1 } & { y }_{ 3 }-{ y }_{ 1 } & { z }_{ 3 }-{ z }_{ 1 } \end{vmatrix}=0\)
i.e.,  \(\begin{vmatrix} x-2 & y-2 & z-1 \\ 3-2 & 0-2 & 1-1 \\ 4-2 & -1-2 & 0-1 \end{vmatrix}=0\)
\(\Rightarrow\)   \(\begin{vmatrix} x-2 & y-2 & z-1 \\ 1 & -2 & 0 \\ 2 & -3 & -1 \end{vmatrix}=0\)
\(\Rightarrow\)  ( x - 2 )( 2 - 0 ) - ( y - 2 ) ( - 1 - 0 )+ ( z - 1 ) ( - 3 + 4 ) = 0
\(\Rightarrow\) 2x + y + z  -7 = 0 ..(i)
When the line crosses the plane, M must satisfy the plane (i).
So, we have
\(2(-\lambda+3)+(\lambda-4)+(6\lambda-5)-7=0\)
\(\Rightarrow\)  \(\lambda=2\)
Hence, required co-ordinates of the point of intersection is M (1, - 2, 7).

\(\therefore\) Co-ordinates of Pare (1, - 2, 7).
Now, for ratio:
Let P divides AB in the ratio k : 1.
\(\therefore\) Using section formula
\(1={{2k+3}\over{k+1}}\)
\(\Rightarrow\) 2k + 3 = k + 1
\(\therefore\)  k = -2
Ratio is - 2 : 1 or 2 : 1 externally.

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).

Line through A(2, -1, 2) and B(5, 3, 4) is  \(\frac { x-2 }{ 3 } =\frac { y+1 }{ 4 } =\frac { z-2 }{ 2 } =\lambda \)
General point on the line is \(\left( 3\lambda +2,\ 4\lambda -1,\ 2\lambda +2 \right) \)
\(\therefore 3\lambda +2-\left( 4\lambda -1 \right) +2\lambda +2=5\Rightarrow \lambda =0\)
Point of intersection is (2, -1, 2)
\(d=\sqrt { \left( 3 \right) ^{ 2 }+{ \left( 4 \right) }^{ 2 }+{ \left( 12 \right) }^{ 2 } } =\sqrt { 169 } =13\)

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

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
\(cos\quad \theta =\frac { \vec { n } _{ 1 }.\vec { n } _{ 2 } }{ \left| \vec { n } _{ 1 } \right| \left| \vec { n } _{ 2 } \right| } \)
\(=\frac { \left( 2\hat { i } +3\hat { j } -\hat { k } \right) \left( \hat { i } +\hat { j } -2\hat { k } \right) }{ \left| \left( 2\hat { i } +3\hat { j } -\hat { k } \right) \right| \left| \left( \hat { i } +\hat { j } -2\hat { k } \right) \right| } \)
\(=\frac { 2+3+2 }{ \left( \sqrt { { 2 }^{ 2 }+{ 3 }^{ 2 }+{ 1 }^{ 2 } } \right) \left( \sqrt { { 1 }^{ 2 }+{ 1 }^{ 2 }+{ 2 }^{ 2 } } \right) } =\frac { 7 }{ 2\sqrt { 21 } } \)