Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

The equation of the line passing through the point $P(1,-5,9)$ and parallel to the line $x=y=z$ (which has direction ratios $1,1,1$) is:

$\frac{x-1}{1}=\frac{y+5}{1}=\frac{z-9}{1}=r$

Any point on this line is of the form $Q(r+1,r-5,r+9)$.

Since the point $Q$ lies on the plane $x-y+z=5$, we substitute its coordinates into the equation of the plane:

$(r+1)-(r-5)+(r+9)=5$

$r+1-r+5+r+9=5$

$r+15=5$

$r=-10$

Substituting $r=-10$, the coordinates of the point of intersection $Q$ are $(-9,-15,-1)$.

The required distance is the distance between $P(1,-5,9)$ and $Q(-9,-15,-1)$:

$PQ=\sqrt{(-9-1{)}^2+(-15-(-5){)}^2+(-1-9{)}^2}$

$PQ=\sqrt{(-10{)}^2+(-10{)}^2+(-10{)}^2}$

$PQ=\sqrt{100+100+100}=\sqrt{300}=10\sqrt{3}$ units.

Answer: $10\sqrt{3}$