Question

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

$3 \sqrt{14}$

Answer 2

eq. of line PM 1x−2​=2y+1​=−1z−3​=λ
any point on line=(λ+2,2λ−1,−λ+3)
for point ' m ’ (λ+2)+2(2λ−1)−(3−λ)=0
λ=$\frac{1}{2}$
Point m($\frac{1}{2}+2,2\times\frac{1}{2},\frac{-1}{2}+3$)
=($\frac{5}{2},0,\frac{5}{2}$​)
For Image Q(α,β,γ)
​$\frac{\alpha + 2}{2}=\frac{5}{2},\frac{\beta-1}{2}=0$
$\frac{\gamma +3}{2}=\frac{5}{2}$
Q:(3,1,2)
$d=|\frac{3(3)+2(1)+2+29}{\sqrt{3^2+2^2+1^2}}|$

$d=\frac{42}{\sqrt{14}}=3\sqrt{14}$