Let the image of the point P(1, 6, a) in the line L: (x)/(1) =...

Let the foot of the perpendicular from $P(1, 6, a)$ to the line $L$ be $M$. Since the image of $P$ is $P'\left(\dfrac{a}{3}, 0, a+c\right)$, $M is the midpoint of$P$and$P'$.<br><br>$M = \left(\dfrac{1 + \dfrac{a}{3}}{2}, \dfrac{6 + 0}{2}, \dfrac{a + a + c}{2}\right) = \left(\dfrac{a+3}{6}, 3, a + \dfrac{c}{2}\right)$<br><br>Since$M$lies on the line$L: \dfrac{x}{1} = \dfrac{y-1}{2} = \dfrac{z-a+1}{b}$, we substitute its coordinates into the equation of the line:<br><br>$\dfrac{\dfrac{a+3}{6}}{1} = \dfrac{3-1}{2} = \dfrac{a + \dfrac{c}{2} - a + 1}{b}$<br><br>$\dfrac{a+3}{6} = 1 = \dfrac{\dfrac{c}{2} + 1}{b}$<br><br>From$\dfrac{a+3}{6} = 1$, we get$a = 3$.<br><br>From$\dfrac{\dfrac{c}{2} + 1}{b} = 1$, we get$2b = c + 2 \Rightarrow c = 2b - 2$.<br><br>The vector$\overrightarrow{PP'}$is perpendicular to the direction vector of the line$L$, which is$\vec{v} = \hat{i} + 2\hat{j} + b\hat{k}$.<br><br>$\overrightarrow{PP'} = \left(\dfrac{a}{3} - 1\right)\hat{i} + (0 - 6)\hat{j} + (a+c - a)\hat{k} = \left(\dfrac{3}{3} - 1\right)\hat{i} - 6\hat{j} + c\hat{k} = -6\hat{j} + c\hat{k}$<br><br>Taking the dot product$\overrightarrow{PP'} \cdot \vec{v} = 0$:<br><br>$0(1) - 6(2) + c(b) = 0 \Rightarrow bc = 12$<br><br>Substituting$c = 2b - 2$into$bc = 12$:<br><br>$b(2b - 2) = 12 \Rightarrow 2b^2 - 2b - 12 = 0 \Rightarrow b^2 - b - 6 = 0$<br><br>$(b-3)(b+2) = 0$<br><br>Since$b > 0$, we have$b = 3$. Then$c = 2(3) - 2 = 4$.<br><br>The coordinates of$M$are \left(\dfrac{3+3}{6}, 3, 3 + \dfrac{4}{2}\right) = (1, 3, 5)$.

The equation of the line $L$ is $\dfrac{x}{1} = \dfrac{y-1}{2} = \dfrac{z-2}{3}$. Any point $S$ on the line can be written as $(t, 2t+1, 3t+2)$.

The distance $SM$ is given as $2\sqrt{14}$.

$SM^2 = (t-1)^2 + (2t+1-3)^2 + (3t+2-5)^2 = 56$

$(t-1)^2 + 4(t-1)^2 + 9(t-1)^2 = 56$

$14(t-1)^2 = 56 \Rightarrow (t-1)^2 = 4 \Rightarrow t - 1 = \pm 2$

$t = 3$ or $t = -1$

Since $S(\alpha, \beta, \gamma) = (t, 2t+1, 3t+2)$ and $\alpha > 0$, we must have $t = 3$.

Thus, $S = (3, 7, 11)$.

$\alpha + \beta + \gamma = 3 + 7 + 11 = 21$