JEE Mains · Maths · STD 12 - 11. three dimension geometry
If the distances of the point (1, 2, a) from the line \(\frac{x-1}{1}=\frac{y}{2}=\frac{z-1}{1}\) along the lines
\(L_{1}:\frac{x-1}{3}=\frac{y-2}{4}=\frac{z-a}{b}\) and
\(L_{2}:\frac{x-1}{1}=\frac{y-2}{4}=\frac{z-a}{c}\) are equal,
then \(a+b+c\) is equal to
- A 7
- B 5
- C 6
- D 4
Answer & Solution
Correct Answer
(A) 7
Step-by-step Solution
Detailed explanation
\begin{aligned} & \mathrm{L}: \frac{\mathrm{x}-1}{1}=\frac{\mathrm{y}}{2}=\frac{\mathrm{z}-1}{1} \\ & \mathrm{~L}_1: \frac{\mathrm{x}-1}{3}=\frac{\mathrm{y}-2}{4}=\frac{\mathrm{z}-\mathrm{a}}{\mathrm{b}}=\lambda \\ & \mathrm{L}_2:…
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