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

The equation of a plane containing the line 

$\frac{x-3}{1}=\frac{y-6}{5}=\frac{z-4}{4}$ is

$a(x-3)+b(y-6)+c(z-4)=0 \cdots \cdots \cdots \cdots \cdots \left(i\right)$

where, $a+5 b+4 c=0\dots \dots \dots \dots \left(ii\right)$

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

$a\left(0\right)+b(-4)+c(-4)=0$

Solving (ij-and (iii), we get 

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

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