The plane containing the point \((3,2,0)\) and the line \(\frac{x-3}{1}=\frac{y-6}{5}=\frac{z-4}{4}\) is

Given that point \((3,2,0)\) lies on plane and line \(\frac{x-3}{1}=\frac{y-6}{5}=\frac{z-4}{4}\) also lies on the plane.
\(\because\) Plane contains points \((3,2,0)\) and \((3,6,4)\)
The DR's of line joining these two points is \((0,4,4)\).
Therefore, DR's of normal to plane is
\((\hat{\mathbf{i}}+5 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}) \times(4 \hat{\mathbf{j}}+4 \hat{\mathbf{k}})=4 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}+4 \hat{\mathbf{k}}\)
Therefore, the equation of plane is
\(4 x-4 y+4 z=k\)
\(\because\) Point \((3,2,0)\) lies on planes. So, we get
\(\therefore \quad k=4 \times 3-4 \times 2+4 \times 0=4\)
Hence, the equation of plane is \(x-y+z=1\).