Question

The line $\frac{x-3}{2}=\frac{y-4}{3}=\frac{z-5}{4}$ lies in the plane 4x+4y-kz-d =0.The values of k and d are respectively

• A. 4,8
• B. -5,-3
• C. 5,3
• D. -4-8

Answer 1

The correct option is C 5,3

Since the line lies on the given plane, therefore any point on the line will satisfy the plane i.e., the points (3, 4, 5) and (5, 7, 9) will lie on the plane.
Hence k=5 and d=3