The 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

The required plane is $\left|\begin{array}{ccc}x-3& y-6& z-4\\ 3-3& 2-6& 0-4\\ 1& 5& 4\end{array}\right|=0$

$\begin{array}{rc}\left|\begin{array}{ccc}\mathrm{x}-3& \mathrm{y}-\mathrm{z}-2& \mathrm{z}-4\\ 0& 0& -4\\ 1& 1& 4\end{array}\right|=0& \\ & \text{ (Operating }\left.{\mathrm{C}}_2\to {\mathrm{C}}_2-{\mathrm{C}}_3\right)\end{array}$

$\begin{array}{l}\mathrm{or} 4(\mathrm{x}-3-\mathrm{y}+\mathrm{z}+2)=0\\ \mathrm{or} \mathrm{x}-\mathrm{y}+\mathrm{z}=1\end{array}$