If A equiv (x, 4, -1),B equiv (3, x, -5),and | vedclass

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(B) The centroid $G$ 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 $G = \left(\frac{x_1+x_2+x_3

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$1$

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(B) The centroid $G$ 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 $G = \left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3}\right)$.Given $A(x, 4, -1)$,$B(3, x, -5)$,$C(2, -2, 3)$,and $G(2, 1, -1)$.Equating the $x$-coordinates: $\frac{x+3+2}{3} = 2$.$x + 5 = 6$.$x = 1$.Checking with $y$-coordinates: $\frac{4+x-2}{3} = 1$ $\Rightarrow 2+x = 3$ $\Rightarrow x = 1$.Thus,the value of $x$ is $1$.

If $A \equiv (x, 4, -1)$,$B \equiv (3, x, -5)$,and $C \equiv (2, -2, 3)$ are the vertices and $G \equiv (2, 1, -1)$ is the centroid of the triangle $ABC$,then the value of $x$ is

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