Question

Sol. [Key Concept: $Q$ is the midpoint of $P$ and its image $P'$; the direction vector of $L$ is perpendicular to $PQ$, fixing $a$; write $S = (k,2k + 1,3k + 2)$ and use $SQ = 2\sqrt{14}$ to get $\alpha + \beta + \gamma$.]

Let Q be the point $\bot^{r}$ from $P(1,6,a)$ to the line

$$L:\frac{x}{1} = \frac{y - 1}{2} = \frac{z - a + 1}{\text{ }b} $$The image of point P in line L is $P^{'}\left( \frac{a}{3},0,a + c \right)$

∵ Q is the mid point ${PP}^{'}$ :

$$Q \equiv \left( \frac{1 + \frac{a}{3}}{2},\frac{6 + 0}{2},\frac{a + a + c}{2} \right) \equiv \left( \frac{3 + a}{6},3,\frac{2a + c}{2} \right) $$comparing $x$ and $y,\frac{3 + a}{6} = 1 \Rightarrow a = 3$

$$\therefore Q \equiv \left( 1,3,\frac{6 + c}{2} \right) $$Direction vector of $L$ is $\overrightarrow{v} = (1,2,b)$

vector PQ is $\bot^{r}$ to $\overrightarrow{V}$

$${\therefore\overrightarrow{PQ} = \left( 1 - 1,3 - 6,\frac{6 + c}{2} - 3 \right) }{= \left( 0, - 3,\frac{c}{2} \right) }{\overrightarrow{PQ} \cdot \overrightarrow{v} = 0 }{(0).(1) + ( - 3)(2) + \left( \frac{c}{2} \right)(b) = 0 }{bc = 12 }$$Now comparing z with $Q\left( 1,3,\frac{6 + c}{2} \right)$

$${\frac{3 - 1}{2} = \frac{\frac{6 + c}{2} - 3 + 1}{b} }{2\text{ }b - c = 2 }$$solving $bc = 12$ and $c = 2b - 2$

$${b = 3, - 2 }{b = 3\ (\text{ }b > 0) }{c = 4 }{\therefore Q(1,3,5) }$$Now, any point S on the line L can be written as $(k,2k + 1,3k + 2)$

$${SQ = \sqrt{(k - 1)^{2} + (2k - 2)^{2} + (3k - 3)^{2}} = 2\sqrt{14} }{k = 3, - 1 }$$for $k = 3,\text{ }S \equiv (3,7,11)$

$$\therefore\alpha + \beta + \gamma = 21 $$