Question

Let $L_1:\frac{x-1}{3}=\frac{y-2}{1}=\frac{z-3}{-3}$ be a line and $P:4x+3y+5z-50=0$ 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 $ax+by+5z+d=0,$ then the value of $a+b+d$ is

Answer 1

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.

$\therefore$ Direction ratios of normal
$=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 3& 1& -3\\ 4& 3& 5\end{array}\right|$

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