Paragraph: Let Q be the image of the point P(-2,1,-5) in the plane...

Let $Q \left(x_1, y_1, z_1\right)$ be the image of the point $P$. The direction ratios of $P Q$ are $3,-2,2$....(i) The Equation of line $P Q$ is $\frac{x+2}{3}=\frac{y-1}{-2}=\frac{z+5}{2}=$ r$Coordinates of any point on the line$P Q$is 3 r-2,-2 r+ 1 and 2 r-5$. Let $Q(3 r-2,-2 r+1,2 r-5)$ be such a point. Let $L$ be the mid point of $P Q, L= \left(\frac{3 r}{2}-2,1-r, r-5\right)$ Since $L$ lies on the plane $3 x-2 y+2 z+1=0$ So, $3\left(\frac{3 r}{2}-2\right)$-2(1-r)+2(r-5)+1=0\Rightarrow $\frac{17}{2}$ r-17=0 \Rightarrow r=2$So, coordinates of$Q$are$(3 \times2-2,-2 \times2+1,2 \times2-5)=(4,-3,-1)$...(ii) Also the mid point of$P Q$is$L=\left(\frac{3 \times 2}{2}-2,1-2,2-5\right)=$(1,-1,-3)$...(iii) $$\begin{aligned} & \therefore P Q=\sqrt{(-2-4)^2+(1+3)^2+(-5+1)^2}=\sqrt{68}\\ & \Rightarrow P Q=2 \sqrt{17} >8 \end{aligned}$$\therefore$ Option (4) is correct.