If G(4, 3, 3) is the centroid of the triangle | vedclass

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(D) The centroid $G(x, y, z)$ of a triangle with vertices $A(x_1, y_1, z_1)$,$B(x_2, y_2, z_2)$,and $C(x_3, y_3, z_3)$ is given by $(\frac{x_1+x_2+x_3

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$a=2, b=3, c=1$

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(D) The centroid $G(x, y, z)$ of a triangle with vertices $A(x_1, y_1, z_1)$,$B(x_2, y_2, z_2)$,and $C(x_3, y_3, z_3)$ is given by $(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3})$.Given $G(4, 3, 3)$,$A(a, 3, 1)$,$B(4, 5, b)$,and $C(6, c, 5)$,we have:$\frac{a+4+6}{3} = 4$ $\Rightarrow a+10 = 12$ $\Rightarrow a = 2$.$\frac{3+5+c}{3} = 3$ $\Rightarrow 8+c = 9$ $\Rightarrow c = 1$.$\frac{1+b+5}{3} = 3$ $\Rightarrow 6+b = 9$ $\Rightarrow b = 3$.Thus,$a=2, b=3, c=1$.

If $G(4, 3, 3)$ is the centroid of the triangle $ABC$ whose vertices are $A(a, 3, 1)$,$B(4, 5, b)$,and $C(6, c, 5)$,then the values of $a, b, c$ are:

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