Let λ 1 ,   λ 2 be the values of λ for which the points 5 2 , 1 ,   λ and - 2 …

Given,

${\lambda }_1, {\lambda }_2$ be the values of $\lambda$ for which the points $\left(\frac{5}{2},1, \lambda \right)$ and $\left(-2, 0, 1\right)$ are at equal distance  from the plane $2x+3y-6z+7$,

So, by using distance formula of a point from a plane we get,

$\left|\frac{-4+0-6+7}{\sqrt{2^2+3^2+6^2}}\right|=\left|\frac{5+3-6\lambda +7}{\sqrt{2^2+3^2+6^2}}\right|$

$\Rightarrow \left|\frac{-4+0-6+7}{7}\right|=\left|\frac{5+3-6\lambda +7}{7}\right|$

$\Rightarrow \frac{3}{7}=\left|\frac{15-6\lambda }{7}\right|$

$\Rightarrow \lambda =2$ or $3$

$\Rightarrow {\lambda }_1=3, {\lambda }_2=2$ as ${\lambda }_1> {\lambda }_2$

So, $\left({\lambda }_1-{\lambda }_2, {\lambda }_2, {\lambda }_1\right)=(1,2,3)$

Now finding the distance of point $\left(1,2,3\right)$ from the line $\frac{x-5}{1}=\frac{y-1}{2}=\frac{z+7}{2}=\lambda$ we get,

Now form diagram we can see that, $\overrightarrow{PM}$ is perpendicular to given line,

So, by perpendicular condition we get,

$\overrightarrow{PM}·\left(\hat{i}+2\hat{j}+2\hat{k}\right)=0$

$\Rightarrow \left(\lambda +4\right)+2\left(2\lambda -1\right)+2\left(2\lambda -10\right)=0$

$\Rightarrow 9\lambda =18 \text{or} \lambda =2$

Hence, the point $M$ will be $M\left(7,5,-3\right)$

So, the distance $\left|\overrightarrow{PM}\right|=\sqrt{6^2+3^2+6^2}=9$