Equation of the plane passing through (-1, 1, 4) and containing the line $\frac{x-1}{3}=\frac{y-2}{1}=\frac{z}{5}$, is :

22y – 9x + z = 35

Equation of any plane containing the line $\frac{x-1}{3}=\frac{y-2}{1}=\frac{z}{5}$ will be

a(x − 1) + b(y − z) + cz = 0

where, 3a + b + 4c = 0         …. (i)

It is given that plane passes through (–1, 1, 4).

∴ –2a – b + 4c = 0                 …. (ii)

From (i) and (ii), we get

$\frac{a}{-9}=\frac{b}{22}=\frac{c}{1}$

Thus the equation of required plane is,

–9(x – 1) + 22(y – 2) + z = 0

i.e., 22y – 9x + z = 35