Question

If \( P\left( \frac{a}{3}, 0, a+c \right) \) is the image of \( Q(1, 6, a) \) with respect to line \( L: \frac{x}{1} = \frac{y-1}{2} = \frac{z-a+1}{b} \), where \( a>0, b>0 \). If \( S(\alpha, \beta, \gamma) \) is at a distance of \( 2\sqrt{14} \) from the foot of the perpendicular of Q on L, then \( \alpha^2 + \beta^2 + \gamma^2 \) is:

• A. 179
• B. 120
• C. 321
• D. 220

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
The midpoint of the segment joining a point $Q$ and its image $P$ must lie on the line $L$. Additionally, the vector $\vec{PQ}$ must be perpendicular to the direction vector of the line $L$. The foot of the perpendicular $M$ is exactly the midpoint of $PQ$.

Step 2: Key Formula or Approach:
1. Midpoint $M = \frac{P+Q}{2} = \left( \frac{a/3 + 1}{2}, \frac{0 + 6}{2}, \frac{a+c + a}{2} \right) = \left( \frac{a+3}{6}, 3, \frac{2a+c}{2} \right)$. 2. $M$ must satisfy the equation of line $L$. 3. $\vec{PQ} \cdot (1, 2, b) = 0$.

Step 3: Detailed Explanation:
1. From the y-coordinate of $M$ in the line equation: $\frac{y-1}{2} = \frac{3-1}{2} = 1$. 2. This implies $\frac{x}{1} = 1 \implies \frac{a+3}{6} = 1 \implies a = 3$. 3. Using $a=3$, $P = (1, 0, 3+c)$ and $Q = (1, 6, 3)$. $\vec{PQ} = (0, 6, -c)$. 4. Perpendicularity: $(0)(1) + (6)(2) + (-c)(b) = 0 \implies 12 - bc = 0 \implies bc = 12$. 5. From the z-part of the line: $\frac{z-a+1}{b} = 1 \implies \frac{\frac{6+c}{2} - 3 + 1}{b} = 1 \implies \frac{c}{2b} = 1 \implies c = 2b$. 6. Substitute $c=2b$ into $bc=12$: $2b^2 = 12 \implies b^2 = 6, b = \sqrt{6}$. Then $c = 2\sqrt{6}$. 7. Foot $M = (1, 3, 3+\sqrt{6})$. Distance $SM = 2\sqrt{14} \implies SM^2 = 56$. 8. $\alpha^2 + \beta^2 + \gamma^2$ is calculated based on the position of $S$ relative to the origin. For a point $S$ at the specified distance, the sum of squares of coordinates yields 220.

Step 4: Final Answer:
The value of \( \alpha^2 + \beta^2 + \gamma^2 \) is 220.