Question

If the line $\frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}{2}$ lies in the plane $x+3y-\alpha z+7=0,\mathrm{then}\alpha$ = _________________.

Answer 1

$\mathrm{Given}\mathrm{line}\frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}{2}$
Lies in the plane x + 3y – αz + 7 = 0
$$\begin{gathered} \mathrm{Whose}\mathrm{normal}\overrightarrow{n}=\hat{i}+3\hat{j}-\alpha \hat{k}\mathrm{then}\overrightarrow{b}=\left(3\hat{i}-5\hat{j}+2\hat{k}\right)\perp \overrightarrow{n}=\left(\hat{i}+3\hat{j}-\alpha \hat{k}\right) \\ \mathrm{i}.\mathrm{e}.3\left(1\right)-5\left(3\right)+2\left(-\alpha \right)=0 \\ \mathrm{i}.\mathrm{e}.3-15-2\alpha =0 \\ \mathrm{i}.\mathrm{e}.-12-2\alpha =0 \\ \mathrm{i}.\mathrm{e}.2\alpha =-12 \\ \mathrm{i}.\mathrm{e}.\alpha =-6 \end{gathered}$$