Let the plane containing the line of intersection of the planes

P₁ : $x+(\lambda+4)y+z=1$ and

P₂ : $2x+y+z=2$

pass through the points (0, 1, 0) and (1, 0, 1). Then the distance of

the point (2$\lambda,\lambda,-\lambda$) from the plane P₂ is :

Equation of plane passing through point of intersection of $\mathrm{P} 1$ and $\mathrm{P} 2$<br/><br/> $$ \begin{aligned} & \mathrm{P}_1+\mathrm{kP}_2 = 0 \\\\ & (\mathrm{x}+(\lambda+4) \mathrm{y}+\mathrm{z}-1)+\mathrm{k}(2 \mathrm{x}+\mathrm{y}+\mathrm{z}-2)=0 \end{aligned} $$<br/><br/> Passing through $(0,1,0)$ and $(1,0,1)$<br/><br/> $$ \begin{aligned} & (\lambda+4-1)+\mathrm{k}(1-2)=0 \\\\ & (\lambda+3)-\mathrm{k}=0\quad...(1) \end{aligned} $$<br/><br/> Also passing $(1,0,1)$<br/><br/> $$ \begin{aligned} & (1+1-1)+\mathrm{k}(2+1-2)=0 \\\\ & 1+\mathrm{k}=0 \\\\ & \mathrm{k}=-1 \end{aligned} $$<br/><br/> put in (1)<br/><br/> $$ \begin{aligned} & \lambda+3+1=0 \\\\ & \lambda=-4 \end{aligned} $$<br/><br/> Then point $(2 \lambda, \lambda,-\lambda)$<br/><br/> $$ \begin{aligned} & (-8,-4,4) \\\\ & d=\left|\frac{-16-4+4-2}{\sqrt{6}}\right| \\\\ & d=\frac{18}{\sqrt{6}} \times \frac{\sqrt{6}}{\sqrt{6}}=3 \sqrt{6} \end{aligned} $$