A plane which passes through the point (3, 2, 0) and the line $\frac{x-3}{1}=\frac{y-6}{5}=\frac{z-4}{4}$, is

$x- y + z = 1 $

The equation of a plane containing the line $\frac{x-3}{1}=\frac{y-6}{5}=\frac{z-4}{4}$ is

$a(x-3) + b(y-6) + c(z-4) = 0 $ .....(i)

where, $a + 5b + 4c = 0 $ ..........(ii)

This plane will pass through (3, 2, 0) if

$a(0)+ b(-4) + c(-4) = 0 $ .........(iii)

Solving (i) and (iii), we get

$\frac{a}{-20+16}=\frac{b}{0+4} = \frac{c}{-4+0}⇒ \frac{a}{1}=\frac{b}{-1}=\frac{c}{1}$

Putting the values of a, b, c in (i), we obtain that the equation of the planes is x - y + z -1 = 0.