Question

If the line $\frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}{2}$ lies in the plane $x+3y-\alpha z+\beta =0$. Then $(\alpha ,\beta )$ equals

• A. $(–6,7)$
• B. $(6,–17)$
• C. $(–5,5)$
• D. $(5,–15)$

Answer 1

The correct option is A

$(–6,7)$

The normal to the plane is perpendicular to the line.

Normal to the given plane is $\hat{i}+3\hat{j}-\alpha \hat{k}=0$

So, $(3\hat{i}-5\hat{j}+2\hat{k}).(\hat{i}+3\hat{j}-\alpha \hat{k})=0$

$\Rightarrow 3-15-2\alpha =0\Rightarrow \alpha =-6$

Also, as the line lies in the plane, the point $(2,1,–2)$ also lies in the plane.

$\Rightarrow 1\left(2\right)+3\left(1\right)-(-6)(-2)+\beta =0\Rightarrow \beta =7$

$\Rightarrow (\alpha ,\beta )=(-6,7)$