JEE Mains · Maths · STD 12 - 11. three dimension geometry
The lines \(\frac{x-2}{2}=\frac{y}{-2}=\frac{z-7}{16}\) and \(\frac{x+3}{4}=\frac{y+2}{3}=\frac{z+2}{1}\) intersect at the point \(P\). If the distance of \(P\) from the line \(\frac{x+1}{2}=\frac{y-1}{3}=\frac{z-1}{1}\) is \(l\), then \(14 l^2\) is equal to .................
- A \(108\)
- B \(107\)
- C \(109\)
- D \(100\)
Answer & Solution
Correct Answer
(A) \(108\)
Step-by-step Solution
Detailed explanation
\( \frac{\mathrm{x}-2}{1}=\frac{\mathrm{y}}{-1}=\frac{\mathrm{z}-7}{8}=\lambda \) \( \frac{\mathrm{x}+3}{4}=\frac{\mathrm{y}+2}{3}=\frac{\mathrm{z}+2}{1}=\mathrm{k} \) \( \Rightarrow \lambda+2=4 \mathrm{k}-3 \) \( -\lambda=3 \mathrm{k}-2 \)…
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