If left(frac{9}{4}, frac{5}{4}, frac{15}{4}ri | vedclass

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(D) The centroid of a tetrahedron with vertices $(x_1, y_1, z_1), (x_2, y_2, z_2), (x_3, y_3, z_3)$ and $(x_4, y_4, z_4)$ is given by $\left(\frac{x_1

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$a=c=b+1$

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(D) The centroid of a tetrahedron with vertices $(x_1, y_1, z_1), (x_2, y_2, z_2), (x_3, y_3, z_3)$ and $(x_4, y_4, z_4)$ is given by $\left(\frac{x_1+x_2+x_3+x_4}{4}, \frac{y_1+y_2+y_3+y_4}{4}, \frac{z_1+z_2+z_3+z_4}{4}\right)$.Given vertices are $(a, 2, 1), (1, b, 4), (4, 0, c)$ and $(1, 1, 7)$.So, the centroid is $\left(\frac{a+1+4+1}{4}, \frac{2+b+0+1}{4}, \frac{1+4+c+7}{4}\right) = \left(\frac{a+6}{4}, \frac{b+3}{4}, \frac{c+12}{4}\right)$.Equating this to the given centroid $\left(\frac{9}{4}, \frac{5}{4}, \frac{15}{4}\right)$:For $x$-coordinate: $\frac{a+6}{4} = \frac{9}{4} \Rightarrow a+6 = 9 \Rightarrow a = 3$.For $y$-coordinate: $\frac{b+3}{4} = \frac{5}{4} \Rightarrow b+3 = 5 \Rightarrow b = 2$.For $z$-coordinate: $\frac{c+12}{4} = \frac{15}{4} \Rightarrow c+12 = 15 \Rightarrow c = 3$.Thus, $a=3, b=2, c=3$. Comparing these values, we get $a=c=b+1$ (since $3=3=2+1$).

If $\left(\frac{9}{4}, \frac{5}{4}, \frac{15}{4}\right)$ is the centroid of a tetrahedron whose vertices are $(a, 2, 1), (1, b, 4), (4, 0, c)$ and $(1, 1, 7)$, then

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