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 $\left(\alpha ,\beta \right)$ equals

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

Answer 1

Answer: (A)

$\because$ The line $\frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}{2}$ lie in the plane
$x+3y-\alpha z+\beta =0$
$\therefore Pt\left(2,1,-2\right)$ lies on the plane i.e. $2+3+2\alpha +\beta =0$
or $2\alpha +\beta +5=0....\left(i\right)$
Also normal to plane will be perpendicular to line,
$\therefore 3\cdot 1-5\times 3+2\times \left(-\alpha \right)=0\Rightarrow \alpha =-6$
From equation $\left(i\right)$ then, $\beta =7$
$\therefore \left(\alpha ,\beta \right)=\left(-6,7\right)$