MHT CET · Maths · Properties of Triangles
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
- A 15
- B -7
- C 7
- D -5
Answer & Solution
Correct Answer
(C) 7
Step-by-step Solution
Detailed explanation
Centroid of tetrahedron
\(\begin{aligned}
& \equiv\left(\frac{5+a+4+6}{4}, \frac{-7+5-6+c}{4}, \frac{0+3+b+2}{4}\right) \\
\therefore \quad & (4,-3,2) \equiv\left(\frac{15+a}{4}, \frac{-8+c}{4}, \frac{b+5}{4}\right)
\end{aligned}\)
\(\begin{aligned} & \Rightarrow \frac{15+a}{4}=4 \Rightarrow a=1 \\ & \frac{-8+c}{4}=-3 \Rightarrow c=-4 \\ & \frac{b+5}{4}=2 \Rightarrow b=3 \\ \therefore \quad & 2 a+3 b+c=2(1)+3(3)-4=7\end{aligned}\)
\(\begin{aligned}
& \equiv\left(\frac{5+a+4+6}{4}, \frac{-7+5-6+c}{4}, \frac{0+3+b+2}{4}\right) \\
\therefore \quad & (4,-3,2) \equiv\left(\frac{15+a}{4}, \frac{-8+c}{4}, \frac{b+5}{4}\right)
\end{aligned}\)
\(\begin{aligned} & \Rightarrow \frac{15+a}{4}=4 \Rightarrow a=1 \\ & \frac{-8+c}{4}=-3 \Rightarrow c=-4 \\ & \frac{b+5}{4}=2 \Rightarrow b=3 \\ \therefore \quad & 2 a+3 b+c=2(1)+3(3)-4=7\end{aligned}\)
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