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 $(\alpha ,\beta )$ equals :

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

Answer 1

The correct option is A $(-6,7)$

The line is $\frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}{2}$

The direction ratios of the line are $(3,-5,2)$

As the line is in the plane $x+3y-az+\beta =0$,

We have $\left(3\right)\left(1\right)+(-5)\left(3\right)+2(-\alpha )=0$

$\Rightarrow -12-2\alpha =0$

$\therefore \alpha =-6$

Again $(2,1,-2)$ lies on the plane

$\Rightarrow 2+3+2\alpha +\beta =0$

$\Rightarrow \beta =-2\alpha -5=12-5=7$

Hence, $(\alpha ,\beta )$ is $(-6,7)$