We know, if the image of a point P (x₀, y₀, z₀) on a plane Ax + By + Cz + D=0……… (1) is Q (x₁, y₁, z₁) then,

The given plane is, 2x–y + z + 3=0 …………. (2)

Comparing equation (2) with equation (1) we get,

A=2, B= - 1, C=1, D=3

And, here x₀=1, y₀=3, z₀=4

So,

x_1=(2⨯( - 2)) + 1

= - 4 + 1

x_1= - 3,

y_1=(( - 1)⨯( - 2)) + 3

=2 + 3

y_1=5 and

z_1=(1⨯( - 2)) + 4

= - 2 + 4

z_1=2

So, the image of the point (1, 3, 4) in the plane 2x – y + z + 3 = 0 is ( - 3, 5, 2)