Let the plane containing the line of intersection of the planes P _...

Given,

Equation of planes ${P}_{1}: x + \left(\lambda + 4\right)y + z = 1$ and {P}_{2}: 2x + y + z = 2

Now we know that equation of plane passing through the planes ${P}_{1} \text{\text{ and } }$ {P}_{2} is given by ${P}_{1} + k{P}_{2} = 0$

Now given it is passing through the points \left(0, 1, 0\right)$and \left(1, 0, 1\right)$,

So, $x + \left(\lambda + 4\right)y + z - 1 + k\left(12x + y + z - 2\right) = 0$ when passing through \left(0, 1, 0\right)$we get,</p><p>$0 + \left(\lambda + 4\right)1 + 0 - 1 + k\left(12 \times 0 + 1 + 0 - 2\right) = 0$</p><p>$⇒ \lambda + 4 - 1 - k = 0$\</p><p>$⇒ \lambda + 3 - k = 0 . . . . . . . . \left(1\right)$</p><p>And when passing through \left(1, 0, 1\right)$ we get, $k = - 1$

So, from equation \left(1\right)$we get,$\lambda = - 4$</p><p>So, point \left(2\lambda, \lambda, - \lambda\right)$ will be \left(- 8, - 4, 4\right)$</p><p>And distance of point\left(- 8, - 4, 4\right)$ from the plane ${P}_{2}$ will be,

$D = \left|\dfrac{- 2 \times 8 - 1 \times 4 + 1 \times 4 - 2}{\sqrt{{2}^{2} + {1}^{2} + {1}^{2}}}\right| = \dfrac{18}{\sqrt{6}} = 3\sqrt{6}$