Question

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

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

Answer 1

Answer: (A)

Since the line $\frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}{2}$ line in the plane $x+3y+\alpha z+\beta =0$ $\therefore$ P (2, 1, -2) lies on this plane.
$\therefore$ $2+3(1)-\alpha (-2)+\beta$ = 0
$\Rightarrow$ $2\alpha +\beta +5$ = 0 ....(1)
Also normal to the plane is $\perp$ to the given line.
$\therefore$ $3\times 1+3(-5)-\alpha (2)=0\Rightarrow -2\alpha =15-3=12\Rightarrow \alpha =-6$
$\therefore$ ( 1 ) gives $-12+\beta +5$ = 0 $\therefore$ $\beta =12-5=7$
$\therefore$ $(\alpha ,\beta )$ = ( - 6, 7)