<p>The distance of the point \((1, 0, 2)\) from the point of intersection of the line \(\dfrac{x-2}{3} = \dfrac{y+1}{4} = \dfrac{z-2}{12}\) and the plane \(x - y + z = 16\), is</p>
Step-by-Step Solution
Key Concept: Find the intersection point by substituting the parametric form of the line into the plane equation, then calculate the distance using the distance formula.
Step 1: Convert the line to parametric form. Let $\frac{x-2}{3} = \frac{y+1}{4} = \frac{z-2}{12} = t$ Then: $x = 2 + 3t$, $y = -1 + 4t$, $z = 2 + 12t$ Step 2: Substitute into the plane equation $x - y + z = 16$: $(2 + 3t) - (-1 + 4t) + (2 + 12t) = 16$ $2 + 3t + 1 - 4t + 2 + 12t = 16$ $5 + 11t = 16$ $11t = 11 \Rightarrow t = 1$ Step 3: Find the intersection point by substituting $t = 1$: $x = 2 + 3(1) = 5$ $y = -1 + 4(1) = 3$ $z = 2 + 12(1) = 14$ Intersection point: $(5, 3, 14)$ Step 4: Calculate distance from $(1, 0, 2)$ to $(5, 3, 14)$: $d = \sqrt{(5-1)^2 + (3-0)^2 + (14-2)^2}$ $d = \sqrt{16 + 9 + 144} = \sqrt{169} = 13$ ∴ Answer: C
Correct Answer: C