MHT CET · Maths · Three Dimensional Geometry

The distance of the point \((1,6,2)\) from the point of intersection of the line \(\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12}\) and the plane \(x-y+z=16\) is

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

Verified Solution

Answer & Solution

Correct Answer

(C) 13 units

Step-by-step Solution

Detailed explanation

Let \(\frac{x-2}{3}=\frac{y+1}{4}=\frac{z-2}{12}=\lambda\)
\(\therefore \quad\) The co-ordinates of any point on the line are
\(P \equiv(3 \lambda+2,4 \lambda-1,12 \lambda+2)\)
This point lies on the plane \(x-y+z=16\)
\(\therefore \quad 3 \lambda+2-(4 \lambda-1)+12 \lambda+2=16\)
\(\begin{aligned} & \Rightarrow 11 \lambda=11 \\ & \Rightarrow \lambda=1\end{aligned}\)
\(\therefore \quad \mathrm{P} \equiv(5,3,14)\)
Let \(\mathrm{Q} \equiv(1,6,2)\)
\(\therefore \quad \mathrm{PQ}=\sqrt{(1-5)^2+(6-3)^2+(2-14)^2}\)
\(=\sqrt{16+9+144}=13\) units

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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