The equation of the plane passing through the line of intersection of the planes $\overrightarrow r .\left( {\widehat i + \widehat j + \widehat k} \right) = 1$ and $$\overrightarrow r .\left( {2\widehat i + 3\widehat j - \widehat k} \right) + 4 = 0$$ and parallel to the x-axis is :

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