Question

If the image of the point \( P(1, 2, a) \) in the line \[ \frac{x - 6}{3} = \frac{y - 7}{2} = \frac{7 - z}{2} \] is \( Q(5, b, c) \), then \( a^2 + b^2 + c^2 \) is equal to

• A. 293
• B. 298
• C. 264
• D. 283

Hint:

For a point and its image in a line, the line is the perpendicular bisector of the segment joining them. Always use midpoint and direction ratio conditions together.

Answer 1

The Correct Option is D

We are given a point \( P(1, 2, a) \) and its image \( Q(5, b, c) \) in a given line. For a point and its image in a line, the line acts as the perpendicular bisector of the segment joining the point and its image.
Step 1: Write the parametric form of the given line.
From \[ \frac{x - 6}{3} = \frac{y - 7}{2} = \frac{7 - z}{2} = t, \] we get \[ x = 6 + 3t,\quad y = 7 + 2t,\quad z = 7 - 2t. \] Step 2: Find the foot of the perpendicular from \( P \) to the line.
Let the foot of the perpendicular be \[ R(6 + 3t, 7 + 2t, 7 - 2t). \] Since \( R \) is the midpoint of \( P \) and \( Q \), \[ R = \left( \frac{1 + 5}{2}, \frac{2 + b}{2}, \frac{a + c}{2} \right) = (3, \tfrac{2+b}{2}, \tfrac{a+c}{2}). \] Step 3: Equate coordinates to find \( t, a, b, c \).
From the \( x \)-coordinate, \[ 6 + 3t = 3 \Rightarrow t = -1. \] Substitute \( t = -1 \), \[ R = (3, 5, 9). \] Comparing coordinates, \[ \frac{2 + b}{2} = 5 \Rightarrow b = 8, \] \[ \frac{a + c}{2} = 9 \Rightarrow a + c = 18. \] Step 4: Use direction ratios to find \( a \) and \( c \).
Direction ratios of the line are \( (3, 2, -2) \). Vector \( \overrightarrow{PQ} = (4, b - 2, c - a) \) is parallel to the line. Thus, \[ \frac{4}{3} = \frac{6}{2} = \frac{c - a}{-2}. \] From this, \[ c - a = -4. \] Solving \[ a + c = 18,\quad c - a = -4, \] we get \[ a = 11,\quad c = 7. \] Step 5: Calculate \( a^2 + b^2 + c^2 \).
\[ a^2 + b^2 + c^2 = 11^2 + 8^2 + 7^2 = 121 + 64 + 49 = 283. \] Final Answer: \[ \boxed{283} \]