Mathematics - 3D-Coordinate Geometry Question with Solution | TestHub

Let the given point be $P(1,0,-3)$.

The equation of the line passing through $P$ and parallel to the given line $\frac{x-2}{2}=\frac{y+2}{3}=\frac{z-6}{-6}$ is

$\frac{x-1}{2}=\frac{y-0}{3}=\frac{z+3}{-6}=r$

Any point on this line is of the form $Q(2r+1,3r,-6r-3)$.

Since $Q$ lies on the plane $x-y-z=9$, substituting the coordinates of $Q$ in the plane equation gives

$(2r+1)-(3r)-(-6r-3)=9$

$2r+1-3r+6r+3=9$

$5r+4=9\Rightarrow 5r=5\Rightarrow r=1$

Substituting $r=1$, the coordinates of $Q$ are $(3,3,-9)$.

The required distance is the distance between $P$ and $Q$:

$PQ=\sqrt{(3-1{)}^2+(3-0{)}^2+(-9-(-3){)}^2}$

$PQ=\sqrt{2^2+3^2+(-6{)}^2}$

$PQ=\sqrt{4+9+36}=\sqrt{49}=7$ units.

Answer: $7$ units