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

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:

Accepted Answer

Line $PM$: $(x-2)/1=(y+1)/2=(z-3)/-1=\lambda$; point on line $(\lambda+2,\ 2\lambda-1,\ -\lambda+3)$. On the plane:
$$(\lambda+2)+2(2\lambda-1)-(3-\lambda)=0 \implies \lambda=\frac12.$$
$$M=\left(\frac52,0,\frac52\right).$$
Image $Q$: $\dfrac{\alpha+2}{2}=\dfrac52,\ \dfrac{\beta-1}{2}=0,\ \dfrac{\gamma+3}{2}=\dfrac52 \implies Q=(3,1,2).$
$$d=\frac{|3(3)+2(1)+2+29|}{\sqrt{9+4+1}}=\frac{42}{\sqrt{14}}=3\sqrt{14}.$$

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