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

We know that the image of the point $(x_1, y_1, z_1)$ in the plane $ax + by + cz + d= 0 $ is given by

$\frac{x-x_1}{a}=\frac{y-y_1}{b}=\frac{z-z_1}{c}= \frac{-2(ax_1+by_1+cz_1+d)}{a^2+b^2+c^2}$

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

$\frac{x-1}{2}=\frac{y-3}{-1}=\frac{z-4}{1}=\frac{-2(2-3+4-3)}{4+1+1}$

$⇒ \frac{x-1}{2}=\frac{y-3}{-1}=\frac{z-4}{1}= - 2⇒ x = - 3 , y = 5 , z= 2$

Hence, the required point is (-3, 5, 2).