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 .... .
- A \(16\)
- B \(61\)
- C \(65\)
- D \(69\)
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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