Any plane through the line of intersection of planes
x + 2 y + 3 z – 4 = 0
and 2 x + y – z + 5 = 0 is
(x + 2 y + 3 z – 4) + k (2 x + y – z + 5) = 0    ...(1)
i.e. (2 k + 1) x + (k + 2) y + (– k + 3) z + (5 k – 4) = 0
Direction ratios of its normal are 2 k + 1, k + 2, – k + 3
Again consider the plane
5 x + 3 y + 6 z + 8 = 0    ...(2)
Direction ratios of its normal are 5, 3, 6
Since plane (1) is perpendicular to plane (2)
∴ 5 (2 k + 1) + 3 (k + 2) + 6 (– k + 3) = 0
∴ 10 k + 5 + 3 k + 6 – 6 k + 18 = 0

Putting 
               
or 7 (x + 2 y + 3 z – 4) – 29 (2 x + y – z + 5) = 0
or 7 x + 14 y + 21 z – 28 – 58 x – 29 y + 29 z – 145 = 0
or  – 51 x –15 y + 50 z – 173 = 0
or  51 x + 15 y – 50 z + 173 =0
which is required equation of plane.