If the plane $56 x+4 y+9 z=2016$ meets the coordinate axes in $A, B, C$, then the centroid of the $\triangle…
If the plane $56 x+4 y+9 z=2016$ meets the coordinate axes in $A, B, C$, then the centroid of the $\triangle A B C$ is
$(12,168,224)$
$(12,168,112)$
$\left(12,168, \frac{224}{3}\right)$
$\left(12,-168, \frac{224}{3}\right)$
Solution
Given that equation of plane,
$\begin{aligned}
& 56 x+4 y+9 z=2016 \\
& \frac{x}{\frac{2016}{56}}+\frac{y}{\frac{2016}{4}}+\frac{z}{\frac{2016}{9}}=1
\end{aligned}$
Also given, this plane meets the coordinate axes at points $\mathrm{A}, \mathrm{B}$ and $\mathrm{C}$.
Coordinate of $\mathrm{A}=\left(\frac{2016}{56}, 0,0\right)$
Coordinate of $\mathrm{B}=\left(0, \frac{2016}{4}, 0\right)$
Coordinate of $\mathrm{C}=\left(0,0, \frac{2016}{9}\right)$
Now, centroid of $\triangle \mathrm{ABC}$
$\mathrm{G}=\left(\frac{2016}{56 \times 3}, \frac{2016}{4 \times 3}, \frac{2016}{9 \times 3}\right)=\left(12,168, \frac{224}{3}\right)$