Official solution
Equation of planes are
\[\overrightarrow r .\left( {\widehat i + \widehat j + \widehat k} \right) - 1 = 0 \Rightarrow x + y + z - 1 = 0\]
and \[\overrightarrow r .\left( {2\widehat i + 3\widehat j - \widehat k} \right) + 4 = 0 \Rightarrow 2x + 3y - z + 4 = 0\]
equation of planes through line of intersection of these planes is :-
\[(x + y + z - 1) + \lambda (2x + 3y - z + 4) = 0\]
\[ \Rightarrow (1 + 2\lambda )x + (1 + 3\lambda )y + (1 - \lambda )z - 1 + 4\lambda = 0\]
But this plane is parallel to x-axis whose direction are (1, 0, 0)
\[\therefore\] \[(1 + 2\lambda )1 + (1 + 3\lambda )0 + (1 - \lambda )0 = 0\]
\[\lambda = - {1 \over 2}\]
\[\therefore\] Required plane is
\[0x + \left( {1 - {3 \over 2}} \right)y + \left( {1 + {1 \over 2}} \right)z - 1 + 4\left( {{{ - 1} \over 2}} \right) = 0\]
\[ \Rightarrow {{ - y} \over 2} + {3 \over 2}z - 3 = 0\]
\[ \Rightarrow y - 3z + 6 = 0\]
\[ \Rightarrow \overrightarrow r .\left( {\widehat j - 3\widehat k} \right) + 6 = 0\] Ans.