Question

The distance of the point (1, 6, 2) from the point of intersection of the line $(x-2)/3 = (y+1)/4 = (z-2)/12$ and the plane $x - y + z = 16$ is:

• A. 11 units
• B. 12 units
• C. 13 units
• D. 14 units

Hint:

First find the point of intersection, then apply the 3D distance formula.

Answer 1

The Correct Option is C

Step 1: Concept
Intersection of a line and a plane in 3D.

Step 2: Analysis
Let $(x-2)/3 = (y+1)/4 = (z-2)/12 = k$. Point on line: $(3k+2, 4k-1, 12k+2)$.
Substitute into plane $x - y + z = 16$: $(3k+2) - (4k-1) + (12k+2) = 16$.

Step 3: Calculation
$11k + 5 = 16 \implies k = 1$. Intersection point is $(5, 3, 14)$.
Distance from $(1, 6, 2)$: $\sqrt{(5-1)^2 + (3-6)^2 + (14-2)^2} = \sqrt{16 + 9 + 144} = \sqrt{169} = 13$.

Step 4: Conclusion
The distance is 13 units. Final Answer: (C)