Let the lines

$\frac{x-1}{\lambda}=\frac{y-2}{1}=\frac{z-3}{2}$ and

$\frac{x+26}{-2}=\frac{y+18}{3}=\frac{z+28}{\lambda}$ be coplanar

and $\mathrm{P}$ be the plane containing these two lines.

Then which of the following points does NOT lie on P?

<p>${L_1}:{{x - 1} \over \lambda } = {{y - 2} \over 1} = {{z - 3} \over 2}$,</p> <p>through a point ${\overrightarrow a _1} \equiv (1,2,3)$</p> <p>parallel to ${\overrightarrow b _1} \equiv (\lambda ,1,2)$</p> <p>${L_2}:{{x + 26} \over { - 2}} = {{y + 18} \over 3} = {{z + 28} \over \lambda }$</p> <p>through a point ${\overrightarrow a _2} = ( - 26, - 18, - 28)$</p> <p>parallel to ${\overrightarrow b _2} = ( - 2,3,1)$</p> <p>If lines are coplanar then, $$({\overrightarrow a _2} - {\overrightarrow a _1})\,.\,{\overrightarrow b _1} \times {\overrightarrow b _2} = 0$$</p> <p>$$ \Rightarrow \left| {\matrix{ {27} & {20} & {31} \cr \lambda & 1 & 2 \cr { - 2} & 3 & \lambda \cr } } \right| = 0 \Rightarrow \lambda = 3$$</p> <p>Vector normal to the required plane $\overrightarrow n = {\overrightarrow b _1} \times {\overrightarrow b _2}$</p> <p>$$ \Rightarrow \overrightarrow n = \left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 3 & 1 & 2 \cr { - 2} & 3 & 3 \cr } } \right| = - 3\widehat i - 13\widehat j + 11\widehat k$$</p> <p>Equation of plane</p> <p>$\equiv ((x - 1),(y - 2),(z - 3))\,.\,( - 3, - 13,11) = 0$</p> <p>$\Rightarrow 3x + 13y - 11z + 4 = 0$</p> <p>Checking the option gives (0, 4, 5) does not lie on the plane.</p>