Question

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 $P$ be the plane containing these two lines. Then which of the following points does NOT lies on P?

• A. $(0,-2,-2)$
• B. $(-5,0,-1)$
• C. $(3,-1,0)$
• D. $(0,4,5)$

Answer 1

Answer: (D)

Given, $L _1: \frac{ x -1}{\lambda}=\frac{ y -2}{1}=\frac{ z -3}{2}$
and $L_2: \frac{ x +26}{-2}=\frac{ y +18}{3}=\frac{ z +28}{\lambda}$ are coplanar
$\Rightarrow\begin{vmatrix}27 & 20 & 31 \\ \lambda & 1 & 2 \\ -2 & 3 & \lambda\end{vmatrix}=0$
$\Rightarrow \lambda=3$
Now, normal of plane $P$, which contains $L _1$ and $L_2$
$=\begin{vmatrix}\hat{ i } & \hat{ j } & \hat{ k } \\ 3 & 1 & 2 \\ -2 & 3 & 3\end{vmatrix}$
$=-3 \hat{ i }-13 \hat{ j }+11 \hat{ k } $
$ \Rightarrow $Equation of required plane $ P $
$3 x +13 y -11 z +4=0$
$(0,4,5)$ does not lie on plane $P$.