The equation of plane is
2x – y + z + 3 = 0    ...(1)
Direction ratios of normal to the plane are 2,–1, 1.
Let M be foot of perpendicular from P(l, 3, 4) to the plane.
Now PM is a straight line which passes through P( 1. 3. 4) and has direction ratios as 2, – 1, 1
∴ its equation is


Any point M on line is (2 r + 1, – r + 3, r + 4)
∴  M lies on plane (1)
∴ 2 (2 r + 1) – (– r + 3) + (r + 4) + 3 = 0
∴ 4r + 2 + r – 3 + r + 4 + 3 = 0.
∴  6r = –6 ⇒ r = –1
∴ M is (– 2 + 1, 1 + 3, – 1 + 4) i.e. (–1,4, 3)
Let N (α, β,γ) be image of P in the plane (1) so that M is mid-point of PN.