Official solution
P1 : x + y + z -- 6 = 0
P2 : 2x + 3y + z + 5 = 0
Equation of plane which passes through the line of intersection of P1 and P2 is
P1 + \[\lambda \]P2 = 0
\[ \Rightarrow \] (x + y + z -- 6) + \[\lambda (2x + 3y + z + 5) = 0
\[ \Rightarrow \] (1 + 2\[\lambda \])x + (1 + 3\[\lambda \])y + (1 + \[\lambda \]) + (5\[\lambda \] - 6) = 0
As the above plane is perpendicular to xy plane
\[ \therefore \] \[\overrightarrow n .\widehat k = 0\]
\[ \Rightarrow \] \[\left[ {\left( {2 + \lambda } \right)\widehat i + \left( {3 + \lambda } \right)\widehat j + \left( {1 + \lambda } \right)\widehat k} \right].\widehat k\] = 0
\[ \Rightarrow \] 1 + \[\lambda \] = 0
\[ \Rightarrow \] \[\lambda \] = -1
So, equation of plane
- x - 2y - 11 = 0
\[ \Rightarrow \] x + 2y + 11 = 0
Distance of the point (0, 0, 256) from this plane
= \[\left| {{{0 + 0 + 11} \over {\sqrt 5 }}} \right|\] = \[{{11} \over {\sqrt 5 }}\]