Let the plane P pass through the intersection of the planes 2x + 3y...

A plane $P$ pass through the intersection of planes ${P}_{1} and {P}_{2}$ can be written as,

$P ≡ {P}_{1} + \lambda{P}_{2} = 0$

\left(2 + \lambda\right)x + \left(3 + 2\lambda\right)y + \left(3\lambda - 1\right)z - \left(2 + 6\lambda\right) = 0$</p><p>Given,$P ⊥ {P}_{3} ( 2x + y - z + 1 = 0 )$</p><p>$∴ \overset{\to}{n} . \overset{\to}{n}3 = 0$</p><p>$⇒ 2\left(\lambda + 2\right) + \left(2\lambda + 3\right) - \left(3\lambda - 1\right) = 0$</p><p>$⇒ 4 + 2\lambda + 2\lambda + 3 - 3\lambda + 1 = 0$</p><p>$⇒ \lambda = - 8$</p><p>$∴ P ≡ - 6x - 13y - 25z + 46 = 0$</p><p>$6x + 13y + 25z - 46 = 0$</p><p>Since, the distance$d$from$( {x}_{0}, {y}_{0}, {z}_{0} )$to the plane$Ax + By + Cz + D = 0$is,</p><p>$d = \dfrac{\left|A{x}_{0} + B{y}_{0} + C{z}_{0} + D\right|}{\sqrt{{A}^{2} + {B}^{2} + {C}^{2}}}$</p><p>Now the distance of plane$P$from point \left(- 7, 1, 1\right)$,

$d = \dfrac{\left|- 42 + 13 + 25 - 46\right|}{\sqrt{36 + 169 + 625}} = \dfrac{50}{\sqrt{830}}$

$∴ {d}^{2} = \dfrac{50 \times 50}{830} = \dfrac{250}{83}$