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 $(\alpha ,\beta )$ is equal to

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

Answer 1

Answer: (B)

$(-6,7)$

Given line

$\frac{x-2}{3}=\frac{y-1}{-5}=\frac{z+2}{2}$ lies on plane $x+3y-\alpha z+\beta =0$

Then condition to be parallel of line and the plane is

$a_1{a}_2+b_1{b}_2+c_1{c}_2=0$

$3-15-2\alpha =0$

$-2\alpha =12$

$\alpha =-6$

Putting in the equation of the plane

$x+3y+6x+\beta =0$

Since line lie on the plane, therefore, the plane also contains the point of the line from which the line passes

Hence plane contains Point $(2,1,-2)$

$2+3-12+\beta =0$

$\beta =7$

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