Let the plane containing the line of intersection of the planes  $\mathrm{P}1:\mathrm{x}+(\mathrm{\lambda }+4)\mathrm{y}+\mathrm{z}=1$ and  $\mathrm{P}2:2\mathrm{x}+\mathrm{y}+\mathrm{z}=2$ pass through the points $(0,1,0)\text{ and }(1,0,1)$. Then the distance of the point $(2\mathrm{\lambda },\mathrm{\lambda },-\mathrm{\lambda })$ from the plane P2 is

$\begin{array}{l}{\mathrm{P}}_1:\mathrm{x}+(\mathrm{\lambda }+4)\mathrm{y}+\mathrm{z}-1=0\\ {\mathrm{P}}_2:2\mathrm{x}+\mathrm{y}+\mathrm{z}-2=0\end{array}$
Equation of the plane containing the line of intersection of the plane P1 and P2 is
of the from ${\mathrm{P}}_1+{\mathrm{tP}}_2=0$
$\Rightarrow [\mathrm{x}+(\mathrm{\lambda }+4)\mathrm{y}+\mathrm{z}-1]+\mathrm{t}[2\mathrm{x}+\mathrm{y}+\mathrm{z}-2]=0$
Eq. (1) pass through (0, 1, 0) & (1, 0, 1)
$\begin{array}{r}\text{ i.e., }\left[\right(\lambda +4)+t]-1-2t=0\\ \Rightarrow \lambda -t+3=0\dots \dots \dots .\left(2\right)\end{array}$
Also $(1+2\mathrm{t})+(1+\mathrm{t})-1-2\mathrm{t}=0\Rightarrow \mathrm{t}=-1\dots .\left(3\right)$
$\text{ Eq. (2) \& (3) }\mathrm{\lambda }+1+3=0\Rightarrow \mathrm{\lambda }=-4$
Now $(2\mathrm{\lambda },\mathrm{\lambda },-\mathrm{\lambda })=(-8,-4,4)$
Then distance from (-8, -4, 4) to P2 is
$\begin{array}{l}=\frac{\left|{\mathrm{ax}}_1+{\mathrm{by}}_1+{\mathrm{cz}}_1+\mathrm{d}\right|}{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2+{\mathrm{c}}^2}}\\ =\frac{\left|2\right(-8)+(-4)+(4)-2|}{\sqrt{4+1+1}}=\frac{|-16-4+4-2|}{\sqrt{6}}=\frac{18}{\sqrt{6}}=3\sqrt{6}\end{array}$