Question

Let \( d \) be the distance of the point of intersection of the lines \(\frac{x+6}{3} = \frac{y}{2} = \frac{z+1}{1}\) and \(\frac{x-7}{4} = \frac{y-9}{3} = \frac{z-4}{2}\) from the point \((7, 8, 9)\). Then \( d^2 + 6 \) is equal to:

• A. 72
• B. 69
• C. 75
• D. 78

Answer 1

The Correct Option is C

Let the equations of the lines be: 
For Line 1: 
\[ \frac{x + 6}{3} = \frac{y}{2} = \frac{z + 1}{1} = \lambda \] Then \(x = 3\lambda - 6\), \(y = 2\lambda\), \(z = \lambda - 1\). 
For Line 2: 
\[ \frac{x - 7}{4} = \frac{y - 9}{3} = \frac{z - 4}{2} = \mu \] Then \(x = 4\mu + 7\), \(y = 3\mu + 9\), \(z = 2\mu + 4\). 
By equating the coordinates, we get the system of equations: 
\[ \begin{aligned} 3\lambda - 6 &= 4\mu + 7 \quad (1) \\ 2\lambda &= 3\mu + 9 \quad (2) \\ \lambda - 1 &= 2\mu + 4 \quad (3) \end{aligned} \] Solving these equations, we find the values of \(\lambda\) and \(\mu\) at the point of intersection as \(\lambda = 3\) and \(\mu = -1\). Thus, the intersection point is \((3, 6, 2)\). 
The distance \(d\) from the point \((7, 8, 9)\) to \((3, 6, 2)\) is: 
\[ d = \sqrt{(7 - 3)^2 + (8 - 6)^2 + (9 - 2)^2} = \sqrt{16 + 4 + 49} = \sqrt{69} \] Therefore, 
\[ d^2 + 6 = 69 + 6 = 75 \]