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

We have,

Equation of line $PM$ is

$\dfrac{x - 2}{1} = \dfrac{y + 1}{2} = \dfrac{z - 3}{- 1} = \lambda$

Any point on the line is

$M ≡ \left(\lambda + 2, 2\lambda - 1, - \lambda + 3\right)$

Point $M$ lies on the plane, so

\left(\lambda + 2\right) + 2\left(2\lambda - 1\right) - \left(3 - \lambda\right) = 0$</p><p>$⇒ \lambda = \dfrac{1}{2}$</p><p>Point$M ≡ \left(\dfrac{5}{2}, 0, \dfrac{5}{2}\right)$</p><p>For Image$Q\left(\alpha, \beta, \gamma\right)$, we have</p><p>$\dfrac{\alpha + 2}{2} = \dfrac{5}{2}, \dfrac{\beta - 1}{2} = 0$</p><p>$\dfrac{\gamma + 3}{2} = \dfrac{5}{2}$</p><p>So,$Q ≡ \left(3, 1, 2\right)$</p><p>Distance of the point$Q$from the plane$3x + 3y + z + 29 = 0$is</p><p>$d = \left|\dfrac{3\left(3\right) + 2\left(1\right) + 2 + 29}{\sqrt{{3}^{2} + {2}^{2} + {1}^{2}}}\right|$</p><p>$⇒ d = \dfrac{42}{\sqrt{14}} = 3\sqrt{14} units