Question

Let 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, - 17)
• B. (-6, 7)
• C. (5, -15)
• D. (-5, 15)

Answer 1

The correct option is B (-6, 7)

Dr's of line = (3, -5, 2)
Dr's of normal to the plane = (1, 3, - $\alpha$ )
Line is perpendicular to 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
2 + 3 + 6(-2) + $\beta =0\Rightarrow \beta =7$
$\therefore (\alpha ,\beta )=(-6,7)$