Let λ _ 1 , λ _ 2 be the values of λ for which the points ((5)/(2),...

Given,

${\lambda}_{1}, {\lambda}_{2}$ be the values of $\lambda$ for which the points \left(\dfrac{5}{2}, 1, \lambda\right) and \left(- 2, 0, 1\right) are at equal distance from the plane $2x + 3y - 6z + 7,</p><p>So, by using distance formula of$a point from a plane we get,

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

⇒ \left|\dfrac{- 4 + 0 - 6 + 7}{7}\right|$= \left|\dfrac{5 + 3 - 6\lambda + 7}{7}\right|</p><p>$⇒$\dfrac{3}{7}$= \left|\dfrac{15 - 6\lambda}{7}\right|$</p><p>$⇒ \lambda = 2$or 3</p><p>$⇒ {\lambda}_{1} = 3, {\lambda}_{2} = 2$as${\lambda}_{1} > {\lambda}_{2}$</p><p>So, \left({\lambda}_{1} - {\lambda}_{2}, {\lambda}_{2}, {\lambda}_{1}\right)$ = ( 1, 2, 3 )

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

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

So, by perpendicular condition we get,

\overset{\to}{PM} \cdot \left(\overset{^}{i} + 2\overset{^}{j} + 2\overset{^}{k}\right) = 0

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

$⇒ 9\lambda = 18\text{or}\lambda = 2$

Hence, the point $M$ will be $M\left(7, 5, - 3\right)</p><p>So,$ the $distance \left|\overset{\to}{PM}\right| = \sqrt{{6}^{2} + {3}^{2} + {6}^{2}} = 9$