The image of the point (1, 3, 4) in the plane 2x - y + z + 3 = 0, is

Let $(\alpha, \beta, \gamma )$ be the image of the point (1, 3, 4) in the plane 2x - y + z + 3 = 0. Then,

$\frac{\alpha - 1}{2} = \frac{\beta - 3}{-1} = \frac{\gamma - 4}{1}= -\frac{2(2×1-3×1+4×1+3)}{2^2+(-1)^2 + 1^2}$

$⇒ \frac{\alpha - 1}{2} = \frac{\beta - 3}{-1} = \frac{\gamma - 4}{1}= -2 ⇒ \alpha = -3, \beta = 5, \gamma = 2 $

Hence, the image has the coordinates (-3, 5, 2).