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 $\bot$ 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)