Question

Let the line $\frac{x}{1}=\frac{6−y}{2}=\frac{z+8}{5}$ intersect the lines $\frac{x-5}{1}=\frac{y-7}{3}=\frac{z+2}{1}$ and $\frac{x+3}{6}=\frac{3−y}{3}=\frac{z-6}{1}$ at the points A and B respectively. Then the distance of the mid-point of the line segment AB from the plane 2x – 2y + z = 14 is

• A. $\frac{10}{3}$
• B. $\frac{11}{3}$
• C. 4
• D. 3

Answer 1

Answer: (C)

4

Answer 2

The correct option is(C): 4