Question

The equation of the plane passing through the image of the point $P(3,5,7)$ w.r.to the plane $2x+y+z=3$ and having the line $\frac{x-1}{-1}=\frac{y}{2}=\frac{z-2}{-3}$ is:

• A. $x-y-z+1=0$
• B. $3y+2z-4=0$
• C. $19x+5y-3z=13$
• D. $x+5y+3z=7$

Answer 1

The correct option is B $3y+2z-4=0$

Let $P(\alpha ,\beta ,\gamma )$ be the image of the point $Q(3,5,7)=(x_1,y_1,z_1)$ w.r.to the plane $ax+by+cz+d=o=2x+y+z-3$,
We know,
$\frac{\alpha -x_1}{a}=\frac{\beta -y_1}{b}=\frac{\gamma -z_1}{c}=-2\frac{a{x}_1+b{y}_1+c{z}_1+d}{a^2+b^2+c^2}$
$\frac{\alpha -3}{2}=\frac{\beta -5}{1}=\frac{\gamma -7}{1}=-2\frac{2\left(3\right)+1\left(5\right)+1\left(7\right)-3}{4+1+1}=-2\times \frac{15}{6}=-5$
$\alpha =-7,\beta =0,\gamma =2$
Image is $P(-7,0,2)$
Now required plane passes through $P(-7,0,2)$ and contains the line $\frac{x-1}{-1}=\frac{y}{2}=\frac{z-2}{-3}$
So plane passes through $P(-7,0,2)$ and $A(1,0,2)$
D'R's of $AP=(8,0,0)$
$\therefore$ Plane equation is
$\left|\begin{array}{ccc}x-1& y& z-2\\ 8& 0& 0\\ -1& 2& -3\end{array}\right|=0\Rightarrow 3y+2z-4=0$