The equation of plane through (3, 4, 1) is
A(x – 3) + B(y – 4) + C(z – 1) = 0    ...(1)
Since it passes through (0, 1, 0)
∴ A(0 – 3) + B(1 – 4) + C(0 – 1) = 0
or –3A –3B –C = 0
∴ 3A + 3B + C = 0    ..... (2)
The equation of line is

Its direction ratios are 2, 7, 5
Since the line is parallel to plane (1) whose normal has direction ratios A, B, C.
∴ normal to plane (1) is perpendicular to line.
∴ 2A + 7B + 5C = 0    ...(3)
From (2) and (3), we get,
                 

Putting these values of A, B, C in (1), we get,
8 k (x – 3) – 13 k (y – 4) + 15 k (z – 1) = 0
or 8 (x – 3) – 13(y – 4) + 15 (z – 1) = 0
or 8x – 24 – 13 y + 52 + 15 z – 15 = 0
or 8 x – 13 y + 15 z + 13 = 0,
which is required equation of plane.