A plane containing the point (3, 2, 0) and the line $\frac{\mathrm{x}+1}{1}=\frac{\mathrm{y}-2}{5}=\frac{\mathrm{z}-3}{4}$ also contains the point:

$$\begin{gathered} \left|\begin{array}{ccc}\mathrm{x}-3& \mathrm{y}-2& \mathrm{z}-0\\ -4& 0& 3\\ 1& 5& 4\end{array}\right| =0 \Rightarrow (\mathrm{x}-3) (0-15)-(\mathrm{y}-2) (-16-3)+\mathrm{z}(-20-0)=0 \\ -15\mathrm{x}+45+19\mathrm{y}-38-20\mathrm{z}=0 \\ \Rightarrow 15\mathrm{x}-19\mathrm{y}+20\mathrm{z}-7=0 \end{gathered}$$