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. 283
• C. 264
• D. 298

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

Step 1: Coordinates of point N


Given point: \[ A(6,7,7) \] Direction vector: \[ \vec{p} = (3,2,-2) \] Any point \(N\) on the line through \(A\) in the direction of \(\vec{p}\) can be written in parametric form: \[ N = A + \lambda \vec{p} \] \[ N = (6,7,7) + \lambda(3,2,-2) \] \[ N = (3\lambda + 6,\; 2\lambda + 7,\; -2\lambda + 7) \] 

Step 2: Using midpoint condition

Since \(N\) is the midpoint of \(PQ\), \[ P(1,2,a), \quad Q(5,b,c) \] Midpoint formula: \[ N = \left( \frac{1+5}{2}, \frac{2+b}{2}, \frac{a+c}{2} \right) \] \[ N = (3,\; \frac{2+b}{2},\; \frac{a+c}{2}) \] 

Now compare with parametric form of \(N\). From x-coordinate: \[ 3\lambda + 6 = 3 \] \[ 3\lambda = -3 \] \[ \lambda = -1 \] 

Substitute \( \lambda = -1 \) in coordinates of \(N\): \[ N = (3,5,9) \] 

Step 3: Find \(b\) and relation between \(a\) and \(c\)

From y-coordinate: \[ \frac{2+b}{2} = 5 \] \[ 2 + b = 10 \] \[ b = 8 \] From z-coordinate: \[ \frac{a+c}{2} = 9 \] \[ a + c = 18 \] 

Step 4: Use perpendicular condition

From the figure, line \(PN\) is perpendicular to vector \(\vec{p}\). So, \[ \overrightarrow{PN} \cdot \vec{p} = 0 \] First find \(\overrightarrow{PN}\): \[ \overrightarrow{PN} = N - P \] \[ = (3-1,\; 5-2,\; 9-a) \] \[ = (2,\; 3,\; 9-a) \] Now take dot product with \((3,2,-2)\): \[ (2,3,9-a) \cdot (3,2,-2) = 0 \] \[ 2(3) + 3(2) + (9-a)(-2) = 0 \] \[ 6 + 6 -18 + 2a = 0 \] \[ 2a - 6 = 0 \] \[ 2a = 6 \] \[ a = 3 \] 

Step 5: Find \(c\)

Since, \[ a + c = 18 \] \[ 3 + c = 18 \] \[ c = 15 \] 

Step 6: Calculate required value

\[ a^2 + b^2 + c^2 \] \[ = 3^2 + 8^2 + 15^2 \] \[ = 9 + 64 + 225 \] \[ = 298 \] 

Final Answer: \[ \boxed{298} \]