Question

Let the image of the point $P(2,-1,3)$ in the plane $x+2y-z=0$ be $Q$. Then the distance of the plane $3x+2y+z+29=0$ from the point $Q$ is

• A. $2\sqrt{14}$
• B. $\frac{22\sqrt{2}}{7}$
• C. $3\sqrt{14}$
• D. $\frac{24\sqrt{2}}{7}$

Answer 1

Answer: (C)

eq. of line PM $\frac{x-2}{1}=\frac{y+1}{2}=\frac{z-3}{-1}=\lambda$
any point on line$=(\lambda +2,2\lambda -1,-\lambda +3)$
for point ' $m\text{ ’ }(\lambda +2)+2(2\lambda -1)-(3-\lambda )=0$
$\lambda =\frac{1}{2}$
$\text{ Point }m\left(\frac{1}{2}+2,2\times \frac{1}{2}-1,\frac{-1}{2}+3\right)$
$=\left(\frac{5}{2},0,\frac{5}{2}\right)$
For Image $Q(\alpha ,\beta ,\gamma )$
$\frac{\alpha +2}{2}=\frac{5}{2},\frac{\beta -1}{2}=0$
$\frac{\gamma +3}{2}=\frac{5}{2}$
$Q:(3,1,2)$
$d=\left|\frac{3(3)+2(1)+2+29}{\sqrt{3^2+2^2+1^2}}\right|$
$d=\frac{42}{\sqrt{14}}=3\sqrt{14}$