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

Let the direction cosines of the line be $l,m,n$.

Since the line makes equal angles with the coordinate axes, $l=m=n$.

Using the relation $l^2+m^2+n^2=1$ and the fact that the direction cosines are positive, we get $l=m=n=\frac{1}{\sqrt{3}}$.

The equation of the line passing through $P(2,1,2)$ is given by:

$\frac{x-2}{1/\sqrt{3}}=\frac{y-1}{1/\sqrt{3}}=\frac{z-2}{1/\sqrt{3}}=r$

where $r$ represents the distance from point $P$ to any point on the line.

Any point $Q$ on this line can be written as $\left(2+\frac{r}{\sqrt{3}},1+\frac{r}{\sqrt{3}},2+\frac{r}{\sqrt{3}}\right)$.

Since $Q$ lies on the plane $2x+y+z=15$, we substitute its coordinates into the plane equation:

$2\left(2+\frac{r}{\sqrt{3}}\right)+\left(1+\frac{r}{\sqrt{3}}\right)+\left(2+\frac{r}{\sqrt{3}}\right)=15$

$4+\frac{2r}{\sqrt{3}}+1+\frac{r}{\sqrt{3}}+2+\frac{r}{\sqrt{3}}=15$

$7+\frac{4r}{\sqrt{3}}=15$

$\frac{4r}{\sqrt{3}}=8$

$r=2\sqrt{3}$

The length of the line segment $PQ$ is $|r|=2\sqrt{3}$ units.

Answer: $2\sqrt{3}$