Question

Let $L_1: \frac{x-1}{3}=\frac{y-2}{1}=\frac{z-3}{-3}$ be a line and $P: 4 x+3 y+5 z=50$ be a plane. $L_2$ is the line in the plane $P$ and parallel to $L _1$. If equation of the plane containing both the lines $L _1$ and $L _2$ and perpendicular to plane $P$ is $a x+b y+5 z+d=0$ then find the value of $(a+b+d)$.

Answer 1

Here line $L _1$ is parallel to the plane $P$.
$L _2$ is parallel to $L _1$ and lying in the plane $P$
Now, normal of the plane containing $L _1$ and $L _2$
both is $\perp$ to the plane $P$ and the lines.
$\therefore$ Dr's of normal

$\begin{vmatrix}\hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & -3 \\ 4 & 3 & 5\end{vmatrix}=\hat{i}(14)-\hat{j}(27)+\hat{k}(5)=14 \hat{i}-27 \hat{j}+5 \hat{k}$
Hence, equation of the required plane is
$ 14( x -1)-27( y -2)+5( z -3)=0 $
$ 14 x -27 y +5 z +25=0$
$\therefore a + b + d =14-27+25=12 $