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

The equation of the given line is $x=y=2z$, which can be rewritten as $\frac{x}{2}=\frac{y}{2}=\frac{z}{1}$. The direction ratios of this line are $(2,2,1)$.

The equation of the line passing through $P(4,5,3)$ and parallel to the given line is:
$\frac{x-4}{2}=\frac{y-5}{2}=\frac{z-3}{1}=r$

A general point on this line is $Q(2r+4,2r+5,r+3)$.

Since the point $Q$ lies on the plane $x+y+z=2$, we substitute its coordinates into the equation of the plane:
$(2r+4)+(2r+5)+(r+3)=2$

$5r+12=2$

$5r=-10\Rightarrow r=-2$

Substituting $r=-2$, the coordinates of the point of intersection $Q$ are:
$Q(2(-2)+4,2(-2)+5,-2+3)=(0,1,1)$

The required distance is the distance between $P(4,5,3)$ and $Q(0,1,1)$:
$PQ=\sqrt{(4-0{)}^2+(5-1{)}^2+(3-1{)}^2}$

$PQ=\sqrt{16+16+4}=\sqrt{36}=6$ units

Answer: $6$