JEE Mains · Maths · STD 12 - 11. three dimension geometry
Let the area of the triangle formed by the lines \(x+2=y-1=z, \frac{x-3}{5}=\frac{y}{-1}=\frac{z-1}{1}\) and \(\frac{x}{-3}=\frac{y-3}{3}=\frac{z-2}{1}\) be \(A\). Then \(A^2\) is equal to ________
- A 55
- B 56
- C 57
- D 58
Answer & Solution
Correct Answer
(B) 56
Step-by-step Solution
Detailed explanation
\begin{aligned} & \mathrm{L}_1: \mathrm{x}+2=\mathrm{y}-1=\mathrm{z}=\ell \\ & \mathrm{L}_2: \frac{\mathrm{x}-3}{5}=\frac{\mathrm{y}}{-1}=\frac{\mathrm{z}-1}{1}=\mathrm{m} \\ & \mathrm{~L}_3: \frac{\mathrm{x}}{-3}=\frac{\mathrm{y}-3}{5}=\frac{\mathrm{z}-2}{1}=\mathrm{n}…
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