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

(–3, 5, 2)

As it is clear from the figure that PQ will be perpendicular to the plane and foot of this perpendicular is mid point of PQ i.e. N.

So, direction ratios of line PQ is 2, –1, 1

⇒ Equation of line PQ = $\frac{x-1}{2}=\frac{y-3}{-1}=\frac{z-4}{1}$ r (say)

Any point on line PQ is (2r + 1, –r + 3, r + 4)

If this point lies on the plane, then

2(2r + 1) – (–r + 3) + (r + 4) + 3 = 0 ⇒ r = –1

∴ coordinate of foot of perpendicular N = (–1, 4, 3)

As N is middle point of PQ

∴ $-1=\frac{1+x_1}{2}, 4=\frac{3+y_1}{2}, 3=\frac{4+z_1}{2}$

⇒ x₁ = –3, y₁ = 5, z₁ = 2

∴ Image of point P (1, 3, 4) is the point Q (–3, 5, 2).