Question

The value of ${}^{\prime }{k}^{\prime }$ such that $\frac{x-3}{3}=\frac{y-1}{-2}=\frac{z-k}{-8}$ lies in the plane $2x-y+z=10,$ is:

• A. $5$
• B. $3$
• C. $-5$
• D. $-3$

Answer 1

Answer: (A)

$5$

The line : $\frac{x-3}{3}=\frac{y-1}{-2}=\frac{z-k}{-8}$

Let $\frac{x-3}{3}=\frac{y-1}{-2}=\frac{z-k}{-8}$ be $\alpha$

$x=3\alpha +3,y=-2\alpha +1,z=-8\alpha +k$

Putting value of $(x,y,z)$ in $(2x-y+z=10)$

$2(3\alpha +3)-(-2\alpha +1)+(-8\alpha +k)=10$

$\Rightarrow 6\alpha +6+2\alpha -1-8\alpha +k=10$

$\Rightarrow 5+k=10$
$\Rightarrow k=5$