The distance of line $3y-2z-1=0=3x-z+4$ from the point $(2,-1,6)$ is

The correct option is A $2\sqrt{6}$
Direction ratio's of given line
$\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 0& 3& -2\\ 3& 0& -1\end{array}\right|=\hat{i}(-3)-\hat{j}\left(6\right)+\hat{k}(-9)$
$=-3\hat{i}-6\hat{j}-9\hat{k}$

Let $z=0\Rightarrow y=\frac{1}{3}\text{and}x=-\frac{4}{3}$
$\therefore$ Line in Cartesian form is
$\frac{x+\frac{4}{3}}{-3}=\frac{y-\frac{1}{3}}{-6}=\frac{z}{-9}$

Let point of shortest distance be $P\left(\lambda \right)$ i.e.
$P(-\lambda -\frac{4}{3},-2\lambda +\frac{1}{3},-3\lambda )\text{and}Q(2,-1,6)$

For shortest distance $\overrightarrow{PQ}\cdot (\hat{i}+2\hat{j}+3\hat{k})=0$
$((\frac{10}{3}+\lambda )\hat{i}+(2\lambda -\frac{4}{3})\hat{j}+(6+3\lambda \left)\hat{k}\right)\cdot (\hat{i}+2\hat{j}+3\hat{k})=0$
$\Rightarrow \lambda =-\frac{4}{3}$
$\therefore P\equiv (0,3,4)$
$\therefore |PQ|=2\sqrt{6}$