Question

Let image of the line $\frac{x-1}{3}=\frac{y-3}{5}=\frac{z-4}{2}$ in the plane $2x-y+z+3=0$ be $L.$ If a plane $7x+py+qz+r=0$ $(p,q,r\in R)$ is such that it contains the line $L$ and perpendicular to the plane $2x-y+z+3=0,$ then the value of $(p+3q+r)$ is

Answer 1

$\frac{x-1}{3}=\frac{y-3}{5}=\frac{z-4}{2}=\lambda$
Point $A$ on the line is of the form
$x=3\lambda +1,y=5\lambda +3$ and $z=2\lambda +4$
If point $A$ lies on the plane $2x-y+z+3=0,$ then
$2(3\lambda +1)-(5\lambda +3)+2\lambda +4+3=0$
$\Rightarrow 3\lambda +6=0$
$\Rightarrow \lambda =-2$
Thererfore, point $A$ is $(-5,-7,0)$

Now, image of the point $P(1,3,4)$ in the plane $2x-y+z+3=0$
$\frac{x-1}{2}=\frac{y-3}{-1}=\frac{z-4}{1}=-\frac{2(2-3+4+3)}{4+1+1}$
$\Rightarrow \frac{x-1}{2}=\frac{y-3}{-1}=\frac{z-4}{1}=-2$
$\therefore$ Image of $P$ is $Q(-3,5,2)$
D.r's of $AQ$ is $<2,12,2>$
Equation of plane is
$\left|\begin{array}{ccc}x+5& y+7& z-0\\ 2& 12& 2\\ 2& -1& 1\end{array}\right|=0$
$\Rightarrow 14(x+5)+2(y+7)-26z=0$
$\Rightarrow 7x+y-13z+42=0$
But it is given $7x+py+qz+r=0$
$\Rightarrow p=1,q=-13$ and $r=42$
$\therefore p+3q+r=1-39+42=4$