Question

Let P be the image of the point $(3,1,7)$ with respect to the plane $x-y+z=3$. Then the equation of the plane passing through P and containing the straight line
$\frac{x}{1}=\frac{y}{2}=\frac{z}{1}$ is

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

Answer 1

The correct option is C $x-4y+7z=0$

We can write the equation of the line passing though P and Q. We know the coordinates of the point Q and the direction ratios of the line PQ is same as the DCs of the normal. So we get
$\frac{x-3}{1}=\frac{y-1}{-1}=\frac{z-7}{1}=k$
$\therefore$ coordinate of the mid point of PQ is $(k+3,-k+1,k+7)$
from the equation of the plane we get $k=-2$
therefore coordinates of $P$ is
$x=-1,y=5,z=3P(-1,5,3)$


Equation of the plane passing through P and having DCs a, b, c is given by
$a(x+1)+b(y-5)+c(z-3)=0$
Since the line with DCs $1,2,1$ is perpendicular to the normal $a,b,c$ we have
$a+2b+c=0...\left(i\right)$
We also know that point (0, 0, 0) lies on this plane, we get
$a-5b-3c=0...\left(ii\right)$
Using $\left(i\right)$ and $\left(ii\right)$, we get the ratios as
$\frac{a}{-1}=\frac{b}{4}=\frac{c}{-7}-(x+1)+4(y-5)-7(z-3)=0-x+4y-7z=0\Rightarrow x-4y+7z=0$