\(\text { Line is } \frac{2\left(x-\frac{1}{2}\right)}{\sqrt{3}}=\frac{y+2}{2}=\frac{z-3}{3} \)
\(\Rightarrow \frac{x-\frac{1}{2}}{\sqrt{3}}=\frac{y+2}{4}=\frac{z-3}{6} \)
\(\text { DR's of line } A B \text { are } \sqrt{3}, 4 \text { and } 6 \text {. }\)
\(\text { DR's of line parallel to } A B \text { are } \sqrt{3}, 4 \text { and } 6 \text {. }\)
\(\text { Dividing by } \sqrt{3+16+36}=\sqrt{55} \text {, we get }\)
Direction cosines are \(\frac { \sqrt { 3 } }{ \sqrt { 55 } } ,\frac { 4 }{ \sqrt { 55 } } ,\frac { 6 }{ \sqrt { 55 } } \).

\(\vec { r } =x\hat { i } +y\hat { j } +z\hat { k } .\) Equation is 3x + 2y + 5z = 7

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 \(\frac { 3 }{ 3 } =\frac { 5 }{ 5 } =\frac { 3 }{ 3 } \)
\(\Rightarrow \)  AB is parallel to BC, B is common.
Hence, A,B,C are collinear.

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

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

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
\(r\left( \frac { 2 }{ 7 } \check { i } -\frac { 3 }{ 7 } \check { j } +\frac { 6 }{ 7 } \check { k } \right) \)

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,
\(.(\overset { \rightarrow }{ b } -\overset { \rightarrow }{ a } )\times (\overset { \rightarrow }{ c } -\overset { \rightarrow }{ a } )=(-4\check { j } -3\check { i } )=\times (2\check { k } -3\check { i } )\)
\(=\left| \begin{matrix} \check { i } & \check { j } & \check { k } \\ -3 & -4 & 0 \\ -3 & 0 & 2 \end{matrix} \right| \)
= \(-8\check { i } +6\check { j } -12\check { k } \)

\(\hat { r } .(\hat { i } -2\hat { j } -\hat { k } )\)
\(\overset { \rightarrow }{ a } =2\hat { i } -\hat { j } -\hat { k } \)

\(\overset { \rightarrow }{ n } =\hat { i } -2\hat { j } -4\hat { k } ,d=9\)
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