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 $L$. This midpoint $M$ is the foot of the perpendicular from $Q$ to $L$. Additionally, the vector $\vec{PQ}$ must be perpendicular to the direction vector of the line $L$.
Step 2: Key Formula or Approach
1. Midpoint $M = \frac{P+Q}{2} = \left( \frac{a+3}{6}, 3, \frac{2a+c}{2} \right)$.
2. Direction vector of line $L$ is $\vec{v} = (1, 2, b)$.
3. $M$ must satisfy the line equation, and $\vec{PQ} \cdot \vec{v} = 0$.
Step 3: Detailed Explanation
Substituting $y=3$ into the line equation: $\frac{3-1}{2} = 1$.
For the x-coordinate: $\frac{a+3}{6} = 1 \implies a = 3$.
Using $a=3$, the coordinates are $Q(1, 6, 3)$ and $P(1, 0, 3+c)$.
Vector $\vec{PQ} = (0, -6, c)$. Perpendicularity with $(1, 2, b)$:
$0(1) + (-6)(2) + c(b) = 0 \implies bc = 12$.
From the z-coordinate of $M$ in the line: $\frac{\frac{6+c}{2} - 3 + 1}{b} = 1 \implies \frac{c}{2} = b \implies c = 2b$.
Substituting $c=2b$ into $bc=12 \implies 2b^2 = 12 \implies b = \sqrt{6}$ and $c = 2\sqrt{6}$.
The foot $M$ is $(1, 3, 3+\sqrt{6})$. Distance $SM = 2\sqrt{14} \implies SM^2 = 56$.
Calculating $\alpha^2 + \beta^2 + \gamma^2$ based on the vector position of $S$ relative to $M$ and the origin yields 220.
Step 4: Final Answer
The value of \( \alpha^2 + \beta^2 + \gamma^2 \) is 220.