Question

A plane passes through the point $\left(-2,-2,2\right)$ and contains the line joining the points $\left(1,-1,2\right)$ and $\left(1,1,1\right).$ Then the image of $\left(-7,2,3\right)$ in the plane is

• A. $\left(1,-1,5\right)$
• B. $\left(-5,-4,-2\right)$
• C. $\left(-6,-1,-3\right)$
• D. $\left(\frac{13}{23},\frac{7}{23},\frac{6}{23}\right)$

Answer 1

Answer: (C)

The plane passes through the points $\left(-2,-2,2\right),\left(1,-1,2\right),\left(1,1,1\right)$
Its equation is $\left|\begin{array}{ccc}x+2& y+2& z-2\\ 3& 1& 0\\ 3& 3& -1\end{array}\right|=0$
$\Rightarrow x-3y-6z+8=0$
Image of $\left(-7,2,3\right)$ in the plane is
$\frac{x+7}{1}=\frac{y-2}{-3}=\frac{z-3}{-6}=-2\left(\frac{-7-6-18+8}{1+9+36}\right)$
$\frac{x+7}{1}=\frac{y-2}{-3}=\frac{z-3}{-6}=\left(\frac{23}{23}\right)=1$
$x=-6,y=-1,z=-3$