Competency-Based
Assessment Made Simple.

Step-by-Step Solution

Let the given line be $L: \frac{x-1}{3} = \frac{y-3}{2} = \frac{z-2}{2} = \lambda$. Any point on the line $L$ can be written as $(3\lambda+1, 2\lambda+3, 2\lambda+2)$. Let $M$ be the midpoint of $P(1,3,a)$ and $Q(5,b,c)$. Then $M = \left(\frac{1+5}{2}, \frac{3+b}{2}, \frac{a+c}{2}\right) = \left(3, \frac{3+b}{2}, \frac{a+c}{2}\right)$. Since $M$ lies on the line $L$, we have $\frac{3-1}{3} = \frac{\frac{3+b}{2}-3}{2} = \frac{\frac{a+c}{2}-2}{2}$ $\frac{2}{3} = \frac{3+b-6}{4} = \frac{a+c-4}{4}$ $\frac{2}{3} = \frac{b-3}{4} = \frac{a+c-4}{4}$ From $\frac{2}{3} = \frac{b-3}{4}$, we get $8 = 3b - 9$, so $3b = 17$, and $b = \frac{17}{3}$. From $\frac{2}{3} = \frac{a+c-4}{4}$, we get $8 = 3a + 3c - 12$, so $3a + 3c = 20$, and $a+c = \frac{20}{3}$. The line $PQ$ is perpendicular to the line $L$. The direction ratios of $PQ$ are $(5-1, b-3, c-a) = (4, \frac{17}{3}-3, c-a) = (4, \frac{8}{3}, c-a)$. The direction ratios of $L$ are $(3,2,2)$. Since $PQ \perp L$, we have $3(4) + 2(\frac{8}{3}) + 2(c-a) = 0$. $12 + \frac{16}{3} + 2c - 2a = 0$ $36 + 16 + 6c - 6a = 0$ $52 + 6c - 6a = 0$ $6a - 6c = 52$ $3a - 3c = 26$ We have $a+c = \frac{20}{3}$ and $3a-3c = 26$. $a+c = \frac{20}{3} \implies 3a+3c = 20$ Adding the two equations, we get $6a = 46$, so $a = \frac{23}{3}$. Then $c = \frac{20}{3} - a = \frac{20}{3} - \frac{23}{3} = -\frac{3}{3} = -1$. Now, $a = \frac{23}{3}$, $b = \frac{17}{3}$, $c = -1$. $a^2 + b^2 + c^2 = \left(\frac{23}{3}\right)^2 + \left(\frac{17}{3}\right)^2 + (-1)^2 = \frac{529}{9} + \frac{289}{9} + 1 = \frac{529+289+9}{9} = \frac{827}{9}$.