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JEE Mains · Maths · STD 12 - 11. three dimension geometry

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

  1. A \(16\)
  2. B \(61\)
  3. C \(65\)
  4. D \(69\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(61\)

Step-by-step Solution

Detailed explanation

\(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z+1}{6}=\lambda\) \(x=2 \lambda+1, y=3 \lambda+2, z=6 \lambda-1\) for point of intersection of line \(\&\) plane \(2(2 \lambda+1)-(3 \lambda+2)+(6 \lambda-1)=6\) \(7 \lambda=7 \Rightarrow \lambda=1\) \(\text { point }:(3,5,5)\)…
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