Question

The point of intersection of line $\frac{x-6}{-1}=\frac{y+1}{0}=\frac{z+3}{4}$ and plane $x+y-z=3$ is

• A. $(2,1,0)$
• B. $(7,-1,-7)$
• C. $(1,2,-6)$
• D. $(5,-1,1)$

Answer 1

The correct option is D $(5,-1,1)$

The given line is $\frac{x-6}{-1}=\frac{y+1}{0}=\frac{z+3}{4}=r$(say) ..... $\left(i\right)$

And Plane is $x+y-z=3$ ........ $\left(ii\right)$

$\Rightarrow x=-r+6,y=-1,z=4r-3$

Then, the point $(-r+6,-1,4r-3)$ lies on the line $\left(i\right)$.

It is given that the plane and the line intersects

Thus, the point $(-r+6,-1,4r-3)$ satisfies the plane
$\Rightarrow (-r+6)-1-(4r-3)=3\Rightarrow r=1$
$\therefore$ Required intersection point $=(5,-1,1)$.