A school is preparing the stage for its annual day function. They want to place a hanging mic and a…

The mic is at $M(3, 2, 1)$ and the plain backdrop has equation $2x - y + z + 1 = 0$. Distance from $M$ to the plain backdrop: $d = \frac{|2(3) - 2 + 1 + 1|}{\sqrt{2^2 + (-1)^2 + 1^2}} = \frac{|6|}{\sqrt{4 + 1 + 1}} = \frac{6}{\sqrt{6}} = \sqrt{6}\text{ units}$.

The line through $M(3, 2, 1)$ perpendicular to the backdrop $2x - y + z + 1 = 0$ has direction ratios $(2, -1, 1)$. Any point on this line is $(2k + 3, -k + 2, k + 1)$. The foot of perpendicular on the backdrop satisfies: $2(2k + 3) - (-k + 2) + (k + 1) + 1 = 0 \implies 6k + 6 = 0 \implies k = -1$. So the foot of perpendicular on the backdrop is $F(1, 3, 0)$. Since $M(3, 2, 1)$ is equidistant from $F$ and the hanging light $L(x, y, z)$, $M$ is the midpoint of segment $FL$: $\frac{1 + x}{2} = 3 \implies x = 5$ $\frac{3 + y}{2} = 2 \implies y = 1$ $\frac{0 + z}{2} = 1 \implies z = 2$ Therefore, the coordinates of the position of the hanging light are $(5, 1, 2)$.