MHT CET · Maths · Three Dimensional Geometry
If the plane \(\frac{x}{2}+\frac{y}{3}+\frac{z}{6}=1\) cuts the co - ordinate axes at points \(A, B, C\) respectively, then area of the triangle ABC is
- A \(\sqrt{14}\) sq. units
- B \(3 \sqrt{14}\) sq. units
- C \(\frac{1}{\sqrt{14}}\) sq. units
- D \(3 \sqrt{13}\) sq. units
Answer & Solution
Correct Answer
(B) \(3 \sqrt{14}\) sq. units
Step-by-step Solution
Detailed explanation
\(A=(2,0,0), B=(0,3,0), C=(0,0,6)\) Area \( = \frac{1}{2} \sqrt{(ab)^2 + (bc)^2 + (ca)^2} \)
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