The centroid of tetrahedron with vertices $\mathrm{P}(5,-7,0), \mathrm{Q}(\mathrm{a}, 5,3), \mathrm{R}(4,-6,…

The centroid of tetrahedron with vertices $\mathrm{P}(5,-7,0), \mathrm{Q}(\mathrm{a}, 5,3), \mathrm{R}(4,-6, \mathrm{~b})$ and $\mathrm{S}(6, \mathrm{c}, 2)$ is $(4,-3,2)$, then the value of $2 a+3 b+c$ is equal to
  1. 15
  2. -7
  3. 7
  4. -5

Solution

Centroid of tetrahedron $\begin{aligned} & \equiv\left(\frac{5+a+4+6}{4}, \frac{-7+5-6+c}{4}, \frac{0+3+b+2}{4}\right) \\ \therefore \quad & (4,-3,2) \equiv\left(\frac{15+a}{4}, \frac{-8+c}{4}, \frac{b+5}{4}\right) \end{aligned}$ $\begin{aligned} & \Rightarrow \frac{15+a}{4}=4 \Rightarrow a=1 \\ & \frac{-8+c}{4}=-3 \Rightarrow c=-4 \\ & \frac{b+5}{4}=2 \Rightarrow b=3 \\ \therefore \quad & 2 a+3 b+c=2(1)+3(3)-4=7\end{aligned}$

Asked in: MHT CET 2024 (15 May Shift 1)

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