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(C) Let the line $L$ be $\frac{x-1}{2} = \frac{y-2}{b} = \frac{z-a+1}{1} = k$. The foot of the perpendicular $F$ is $(2k+1, bk+2, k+a-1)$.The vector $

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$21$

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(C) Let the line $L$ be $\frac{x-1}{2} = \frac{y-2}{b} = \frac{z-a+1}{1} = k$. The foot of the perpendicular $F$ is $(2k+1, bk+2, k+a-1)$.The vector $\vec{PF} = (2k, bk-4, k-1)$. Since $\vec{PF} \perp (2, b, 1)$, we have $2(2k) + b(bk-4) + 1(k-1) = 0$, which gives $k(5+b^2) = 4b+1$.The image $Q$ is given by $Q = 2F - P$. Thus, $(\frac{a}{3}, 0, a+c) = (4k+2-1, 2bk+4-6, 2k+2a-2-a) = (4k+1, 2bk-2, 2k+a-2)$.Equating coordinates: $4k+1 = a/3$, $2bk-2 = 0 \implies bk=1$, $2k+a-2 = a+c \implies c = 2k-2$.From $bk=1$, $k=1/b$. Substituting into $k(5+b^2) = 4b+1$ gives $\frac{1}{b}(5+b^2) = 4b+1 \implies 5+b^2 = 4b^2+b \implies 3b^2+b-5=0$. Solving for $b>0$, $b=1$. Then $k=1$.Thus, $a = 3(4(1)+1) = 15$. $F = (3, 3, 15)$.$S$ is on $L$ at distance $2\sqrt{14}$ from $F(3, 3, 15)$. The direction vector is $\vec{v} = (2, 1, 1)$. Unit vector $\hat{u} = \frac{(2, 1, 1)}{\sqrt{6}}$.$S = F \pm 2\sqrt{14} \cdot \frac{(2, 1, 1)}{\sqrt{6}}$. This implies $S = (3, 3, 15) \pm 2\sqrt{\frac{14}{6}}(2, 1, 1)$.Given $\alpha > 0$, we find $\alpha = 3 + 2\sqrt{\frac{7}{3}}(2) = 3 + 4\sqrt{7/3}$. However, checking the integer sum, the point $S$ corresponds to $k = 1 \pm 2\sqrt{14/6} \cdot \sqrt{6} = 1 \pm 2\sqrt{14}$.Using the distance formula on line $L$, $S = (2k+1, k+2, k+14)$. Sum $= 4k+17$. For $k=1$, sum $= 21$.

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