A plane P contains the line $x + 2y + 3z + 1 = 0 = x - y - z - 6$, and is perpendicular to the plane $- 2x + y + z + 8 = 0$. Then which of the following points lies on P?

Equation of plane P can be assumed as<br><br>P : x + 2y + 3z + 1 + $\lambda$ (x $-$ y $-$ z $-$ 6) = 0<br><br>$\Rightarrow$ P : (1 + $\lambda$)x + (2 $-$ $\lambda$)y + (3 $-$ $\lambda$)z + 1 $-$ 6$\lambda$ = 0<br><br>$$ \Rightarrow {\overrightarrow n _1} = (1 + \lambda )\widehat i + (2 - \lambda )\widehat j + (3 - \lambda )\widehat k$$<br><br>$\therefore$ ${\overrightarrow n _1}\,.\,{\overrightarrow n _2} = 0$<br><br>$\Rightarrow$ 2(1 + $\lambda$) $-$ (2 $-$ $\lambda$) $-$ (3 $-$ $\lambda$) = 0<br><br>$\Rightarrow$ 2 + 2$\lambda$ $-$ 2 + $\lambda$ $-$ 3 + $\lambda$ = 0 $\Rightarrow$ $\lambda$ = ${3 \over 4}$<br><br>$\Rightarrow$ $P:{{7x} \over 4} + {5 \over 4}y + {{9z} \over 4} - {{14} \over 4} = 0$<br><br>$\Rightarrow$ 7x + 5y + 9z = 14<br><br>(0, 1, 1) lies on P.