Equations of plane passing through $(4,4,0)$ is given by $a(x-4)+b(y-4)+c(z-0)=0$, where $a, b, c$ are DR's of normal to the plane.
Since this plane is $\perp$ to the given plans, therefore, we get
$2 a+b+2 c=0$
and $3 a+3 b+2 c=0$
By cross-multiplication method
$\frac{a}{2-6} =\frac{-b}{4-6}=\frac{c}{6-3}$
$\Rightarrow \frac{a}{-4} =\frac{b}{2}=\frac{c}{3}$
So, the required equation of plane is
$-4(x-4)+2(y-4)+3(z)=0$
$\Rightarrow -4 x+16+2 y-8+3 z=0$
$\Rightarrow 4 x-2 y-3 z=8$