$\frac{\mathit{x}-3}{1}$ = $\frac{\mathit{y}-5}{-2}$ = $\frac{\mathit{z}-7}{1}$
The vector form of this equation is:
$\overrightarrow{\mathit{r}}$ = $3\hat{\mathit{i}}+5\hat{\mathit{j}}+7\hat{\mathit{k}}$ + λ $\left(\hat{\mathit{i}}-2\hat{\mathit{j}}+\hat{\mathit{k}}\right)$
$\overrightarrow{\mathit{r}}$ = ${\overrightarrow{\mathit{a}}}_1+\mathit{\lambda }\overrightarrow{{\mathit{b}}_1}$ ... (1)
$\frac{\mathit{x}+1}{7}$ = $\frac{\mathit{y}+1}{-6}$ = $\frac{\mathit{z}+1}{1}$
The vector form of this equation is:
$\overrightarrow{\mathit{r}}$ = - $\hat{\mathit{i}}-\hat{\mathit{j}}-\hat{\mathit{k}}$ + λ $\left(7\hat{\mathit{i}}-6\hat{\mathit{j}}+\hat{\mathit{k}}\right)$
$\overrightarrow{\mathit{r}}$ = $\overrightarrow{{\mathit{a}}_2}+\mathit{\lambda }\overrightarrow{{\mathit{b}}_2}$
Therefore, $\overrightarrow{{\mathit{a}}_1}$ = $3\hat{\mathit{i}}+5\hat{\mathit{j}}+7\hat{\mathit{k}}$ , $\overrightarrow{{\mathit{b}}_1}$ = $\hat{\mathit{i}}-2\hat{\mathit{j}}+\hat{\mathit{k}}$ , $\overrightarrow{{\mathit{a}}_2}$ = - $\hat{\mathit{i}}-\hat{\mathit{j}}-\hat{\mathit{k}}$ and $\overrightarrow{{\mathit{b}}_2}$ = $7\hat{\mathit{i}}-6\hat{\mathit{j}}+\hat{\mathit{k}}$
Now, the shortest distance between these two lines is given by:
d = $\left|\frac{\overrightarrow{{\mathit{b}}_1}\times \overrightarrow{{\mathit{b}}_2}.\overrightarrow{{\mathit{a}}_2}-\overrightarrow{{\mathit{a}}_1}}{\left|\overrightarrow{{\mathit{b}}_1}\times \overrightarrow{{\mathit{b}}_1}\right|}\right|$
$\overrightarrow{{\mathit{b}}_1}\times \overrightarrow{{\mathit{b}}_2}$ = $\left|\begin{array}{ccc}\hat{\mathit{i}}& \hat{\mathit{j}}& \hat{\mathit{k}}\\ 1& -& 2& 1\\ 7& -6& 1\end{array}\right|$
= $\hat{\mathit{i}}\left(2+6\right)$ - $\hat{\mathit{j}}\left(1-7\right)$ + $\hat{\mathit{k}}\left(-6+14\right)$
= $4\hat{\mathit{i}}+6\hat{\mathit{j}}+8\hat{\mathit{k}}$
$\left|\overrightarrow{{\mathit{b}}_1}\times \overrightarrow{{\mathit{b}}_2}\right|$ = $\sqrt{4^2+6^2+8^2}$ = $\sqrt{116}$
$\overrightarrow{{\mathit{a}}_2}-\overrightarrow{{\mathit{a}}_1}$ = $-\hat{\mathit{i}}-\hat{\mathit{j}}-\hat{\mathit{k}}$ - $3\hat{\mathit{i}}+5\hat{\mathit{j}}+7\hat{\mathit{k}}$
= $-4\hat{\mathit{i}}-6\hat{\mathit{j}}-8\hat{\mathit{k}}$
∴ d = = $\left|\frac{-16-36-64}{\sqrt{116}}\right|$ = $\left|\frac{-116}{\sqrt{116}}\right|$ = $\sqrt{116}$
OR
Let $\frac{\mathit{x}+2}{3}$ = $\frac{\mathit{y}+1}{2}$ = $\frac{\mathit{z}-3}{2}$ = λ
x = 2 + 3 λ ,y = - 1 + 2 λ ,z = 3 + 2 λ
Therefore, a point on this line is: {(-2+3λ), (-1 + 2λ), (3 + 2λ)}
The distance of the point{(-2+3λ), (-1 + 2λ), (3 + 2λ)} from point (1, 2, 3) = $3\sqrt{2}$
∴ = $3\sqrt{2}$
⇒ - 3 + $3{\mathit{\lambda }}^2$ + (-3) + $2\mathit{\lambda }+2{\mathit{\lambda }}^2$ = 18
⇒ 9 + $9{\mathit{\lambda }}^2$ - 18λ + 9 + $4{\mathit{\lambda }}^2$ - 12λ + $4{\mathit{\lambda }}^2$ = 18
$17{\mathit{\lambda }}^2$ - 30λ = 0
λ = 0 , λ = $\frac{30}{17}$
When λ = $\frac{30}{17}$
x = - 2 + 3λ = - 2 + 3 $\left(\frac{30}{17}\right)$ = - 2 + $\frac{90}{17}$ = $\frac{56}{17}$
y = - 1 + 2λ = - 1 + 2 $\left(\frac{30}{17}\right)$ = - 1 + $\frac{60}{17}$ = $\frac{43}{17}$
z = 3 + 2λ = 3 + 2 $\left(\frac{30}{17}\right)$ = $\frac{51+60}{17}$ = $\frac{111}{17}$
Thus, when λ = $\frac{30}{17}$ , the point is $\left(\frac{56}{17},\frac{43}{17},\frac{111}{17}\right)$ and when λ = 0 , the point is (- 2 , - 1 , 3)