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

The coordinates of any point $Q$ on the given line are $(2r+1,-3r-1,8r-10)$ for some $r\in \mathbb{R}$.

The direction ratios of $PQ$ are $2r,-3r-1,8r-10$.

Since $PQ$ is perpendicular to the given line, we have:

$$2(2r)-3(-3r-1)+8(8r-10)=0$$

$$4r+9r+3+64r-80=0$$

$$77r-77=0$$

$$r=1$$

Thus, the coordinates of $Q$, the foot of the perpendicular from $P$ on the line, are $(3,-4,-2)$.

Let $R(a,b,c)$ be the reflection of $P$ in the given line. Since $Q$ is the midpoint of $PR$, we can write:

$$\frac{a+1}{2}=3,\frac{b+0}{2}=-4,\frac{c+0}{2}=-2$$

Solving these equations, we get:

$$a=5,b=-8,c=-4$$

Therefore, the coordinates of the required point are $(5,-8,-4)$.