MHT CET · Maths · Three Dimensional Geometry
If the plane \(\frac{x}{3}+\frac{y}{2}-\frac{z}{4}=1\) cuts the co-ordinate axes at points \(\mathrm{A}, \mathrm{B}\) and C , then the area of the triangle \(A B C\) is
- A \(\frac{\sqrt{61}}{2}\) sq. units
- B \(2 \sqrt{61}\) sq. units
- C \(\sqrt{61}\) sq. units
- D \(3 \sqrt{61}\) sq. units
Answer & Solution
Correct Answer
(C) \(\sqrt{61}\) sq. units
Step-by-step Solution
Detailed explanation
Intercepts: \(a=3, b=2, c=-4\) Area \( = \frac{1}{2} \sqrt{(ab)^2 + (bc)^2 + (ca)^2} \)
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