Question

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

• A. (-5, -15)
• B. (-5, -15)
• C. (-6,7)
• D. (-6, -7)

Answer 1

The correct option is C (-6,7)

DRs of line = (3, -5, 2)
DR's of normal to the plane = (1, 3, -$\alpha$ )
The line is perpendicular to the normal $\Rightarrow 3\left(1\right)-5\left(3\right)+2(-\alpha )=0\Rightarrow 3-15-2\alpha =0\Rightarrow 2\alpha =-12\Rightarrow \alpha =-6$

Also (2, 1, -2) lies on the plane.
$\therefore 2+3+6(-2)+\beta =0\Rightarrow \beta =7$
$\therefore (\alpha ,\beta )=(-6,7)$