The foot of the point $P(1, -3)$ in the plane $2x + 3y - 4z + 22 = 0$ measured parallel to the line $20x = 5y - 4z$ is point $Q$, then the value of $|PQ|^2$ is
Step-by-Step Solution
Key Concept: Use the direction ratios from the parallel condition to parameterize line $PQ$; find the midpoint on the given plane, then compute distance using the distance formula.
Since distance $PQ$ is measured parallel to the line $20x - 5y = 4a$, we have $\frac{a}{1} = \frac{b}{4}$, so the direction ratios of line $PQ$ are $1, 4, 5$. Point $M$ on line $l$ is $(1 + t, 1 + 4\lambda - 2.5\lambda + 3) = M(\text{say})$. Point $M$ lies on plane $2x + 3y - 4z + 22 = 0$, so $2(1+t) + 3(2+2t) - 4(5+3) + 22 = 0 \Rightarrow -6 + 6 + 0 = 0 \Rightarrow \lambda = 1$, giving $M = (2, 2, 8)$. The midpoint $M$ of $PQ$ gives $|PQ| = 2|PM| = 2\sqrt{(2-1)^2 + (2+2)^2 + (8-3)^2} = 2\sqrt{1+16+25} = 2\sqrt{42}$, so $|PQ|^2 = 4 \times 42 = 168$.
Correct Answer: 168