Given :

Equation of plane : 2x - y + z + 3 = 0

P = (1, 3, 4)

To Find : Image of point P in the plane.

Note :

If two vectors with direction ratios (a₁, a₂, a₃) & (b₁, b₂, b₃) are parallel then

Given equation of the plane is

2x - y + z + 3 = 0

⇒2x - y + z = -3 ………..eq(1)

From eq(1) direction ratios of normal vector of the plane are

(2, -1, 1)

Therefore, equation of normal vector is

Given point is P = (1, 3, 4)

Let, R(a, b, c) be image of point P in the given plane.

Therefore, the power of points P and R in the given plane will be equal and opposite.

⇒2a – b + c + 3 = - (2(1) – 3 + 4 + 3)

⇒2a – b + c + 3 = - 6

⇒2a – b + c = - 9 ………eq(2)

Now,

As are parallel

⇒a = 2k+1, b = -k+3, c = k+4

substituting a, b, c in eq(2)

2(2k+1) - (-k+3) + (k+4) = -9

⇒ 4k + 2 + k – 3 + k + 4 = -9

⇒ 6k = -12

,

Therefore, co-ordinates of the image of P are

R(a, b, c) =