Equation of plane P1 + λP₂ = 0

(x + y + z – 6 ) + λ(2x + 3y + z + 5) = 0

⇒ x(1+2λ) + y(1 + 3λ) +2(1+λ)–6+5λ = 0               

This plane is $\perp$ to xy – plane

$\therefore$  1+ λ = 0

So, equation of plane

–x – 2y –11= 0

⇒ x + 2y + 11= 0

distance of the point (0, 0, 256) from this plane.

$=|\frac{0+0+11}{\sqrt{1+4}}=\frac{11}{\sqrt{5}}|$