JEE Mains · Maths · STD 12 - 11. three dimension geometry
Let \(d\) be the distance of the point of intersection of the lines \(\frac{x+6}{3}=\frac{y}{2}=\frac{z+1}{1} \quad\) and \(\frac{x-7}{4}=\frac{y-9}{3}=\frac{z-4}{2}\) from the point \((7,8,9)\). Then \(\mathrm{d}^2+6\) is equal to :
- A \(72\)
- B \(69\)
- C \(75\)
- D \(78\)
Answer & Solution
Correct Answer
(C) \(75\)
Step-by-step Solution
Detailed explanation
\( \frac{x+6}{3}=\frac{y}{2}=\frac{z+1}{1}=\lambda \) .................(\(1\)) \( x=3 \lambda-6, y=2 \lambda, z=\lambda-1 \) \( \frac{x-7}{4}=\frac{y-9}{3}=\frac{z-4}{2}=\mu \) ........................(\(2\)) \( x=4 \mu+7, y=3 \mu+9, z=2 \mu+4 \)…
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