Distance of Point on Line from Foot of Perpendicular

To find the value of $\alpha + \beta + \gamma$, we proceed step-by-step.

Step 1: Find the parameters $a$, $b$, and $c$

The given point is $P(1, 6, a)$ and its image in the line $L : \frac{x}{1} = \frac{y-1}{2} = \frac{z-a+1}{b}, \quad b > 0$ is $P'\left(\frac{a}{3}, 0, a+c\right)$.

The foot of the perpendicular from $P$ to the line $L$ is the midpoint $M$ of the segment $PP'$. Using the midpoint formula: $M = \left( \frac{1 + \frac{a}{3}}{2}, \frac{6 + 0}{2}, \frac{a + a + c}{2} \right) = \left( \frac{a+3}{6}, 3, a + \frac{c}{2} \right)$

Since $M$ lies on the line $L$, its coordinates must satisfy the equation of $L$: $\frac{\frac{a+3}{6}}{1} = \frac{3-1}{2} = \frac{\left(a + \frac{c}{2}\right) - a + 1}{b}$

Equating the first two ratios: $\frac{a+3}{6} = 1 \implies a + 3 = 6 \implies a = 3$

Equating the second and third ratios: $\frac{\frac{c}{2} + 1}{b} = 1 \implies \frac{c}{2} + 1 = b \implies c + 2 = 2b \quad \text{--- (Equation 1)}$

Next, the vector $\vec{PP'}$ must be perpendicular to the direction vector of the line $L$, which is $\vec{v} = (1, 2, b)$. Since $a = 3$, we have $P(1, 6, 3)$ and $P'(1, 0, 3+c)$. Thus: $\vec{PP'} = P' - P = (1 - 1, 0 - 6, (3+c) - 3) = (0, -6, c)$

Taking the dot product $\vec{PP'} \cdot \vec{v} = 0$: $0(1) + (-6)(2) + c(b) = 0 \implies -12 + bc = 0 \implies bc = 12 \quad \text{--- (Equation 2)}$

From Equation 1, $c = 2b - 2$. Substituting this into Equation 2: $b(2b - 2) = 12 \implies 2b^2 - 2b - 12 = 0 \implies b^2 - b - 6 = 0$ $(b - 3)(b + 2) = 0$

Since $b > 0$, we have $b = 3$. Then, $c = 2(3) - 2 = 4$.

Step 2: Determine the line $L$ and the foot of perpendicular $M$

With $a = 3$, $b = 3$, and $c = 4$:

• Point $P = (1, 6, 3)$
• Midpoint (Foot of perpendicular) $M = \left(1, 3, 5\right)$
• Line equation $L : \frac{x}{1} = \frac{y-1}{2} = \frac{z-2}{3} = t$

Any general point $S(\alpha, \beta, \gamma)$ on the line $L$ can be represented in terms of a parameter $t$ as: $S = (t, 2t + 1, 3t + 2)$ So, $\alpha = t$, $\beta = 2t + 1$, and $\gamma = 3t + 2$.

Since $\alpha > 0$, we must have $t > 0$.

Step 3: Find the coordinates of $S(\alpha, \beta, \gamma)$

Note that the parameter $t$ for the foot of perpendicular $M(1, 3, 5)$ on line $L$ is $t_M = 1$.

The distance between $S$ and $M$ along the line $L$ is given by: $SM = |t - 1| \sqrt{1^2 + 2^2 + 3^2} = |t - 1| \sqrt{14}$

We are given that $SM = 2\sqrt{14}$: $|t - 1| \sqrt{14} = 2\sqrt{14} \implies |t - 1| = 2$

This gives two possible values for $t$: $t - 1 = 2 \implies t = 3 \quad \text{or} \quad t - 1 = -2 \implies t = -1$

Since $\alpha = t > 0$, we select $t = 3$.

Thus, the coordinates of point $S(\alpha, \beta, \gamma)$ are: $\alpha = 3$ $\beta = 2(3) + 1 = 7$ $\gamma = 3(3) + 2 = 11$

Step 4: Calculate $\alpha + \beta + \gamma$

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

Correct Option: C