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Q.
The plane $\frac{x}{2}+\frac{y}{3}+\frac{z}{4}=1$ cuts the coordinate axes in $A, B, C$, then the area of the $\triangle A B C$ is :
Bihar CECEBihar CECE 2003
Solution:
If the plane $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$
cuts the coordinate axes in $A, B$ and $C$, then area of triangle $A B C$
$=\frac{1}{2} \sqrt{a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}}$
Given equation of plane is
$\frac{x}{2}+\frac{y}{3}+\frac{z}{4}=1$
On comparing with $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$
$\Rightarrow a=2,\, b=3,\, c=4$
$\therefore $ Area of $\Delta A B C =\frac{1}{2} \sqrt{a^{2} b^{2}+b^{2} c^{2}+c^{2} a^{2}}$
$=\frac{1}{2} \sqrt{2^{2} \cdot 3^{2}+3^{2} \cdot 4^{2}+4^{2} \cdot 2^{2}}$
$=\frac{1}{2} \sqrt{36+144+64}$
$=\frac{1}{2} \sqrt{244}=\frac{1}{2} \cdot 2 \sqrt{61}$
$=\sqrt{61}$