Question

Let the line $\frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}{2}$ lie in the plane $x+3y-\alpha z+\beta =0$, then $(\alpha ,\beta )$ equals

• A. $(7,-7)$
• B. $(-6,7)$
• C. $(2,-16)$
• D. $(-6,6)$

Answer 1

The correct option is B $(-6,7)$

If the given line lie in the plane $x+3y-\alpha z+\beta =0$
$2+3\times 1-\alpha (-2)+\beta =0$
$2\alpha +\beta =-5$ .........(i)

Drs of normal =$1,3,-\alpha$

Drs of line =$3,-5,2$

line is perpendicular to normal
$\Rightarrow 3-15-2\alpha =0$
$\Rightarrow 2\alpha =-12$
$\beta =-5+12=7$ [From (i)]
$(\alpha ,\beta )\equiv (-6,7)$
Hence, option 'B' is correct.