The centroid $G(x, y, z)$ of a triangle with vertices $P(x_1, y_1, z_1)$, $Q(x_2, y_2, z_2)$, and $R(x_3, y_3, z_3)$ is given by the formula:

$G(x, y, z) = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}, \frac{z_1 + z_2 + z_3}{3} \right)$

In this case:

• $G(a, 2, -1)$ is the centroid,
• $P(1, 3, 2)$, $Q(3, b, -4)$, and $R(5, 1, c)$ are the vertices.

We can now set up equations by equating the coordinates of the centroid:

1. For the $x$-coordinate: $a = \frac{1 + 3 + 5}{3} = \frac{9}{3} = 3$

2. For the $y$-coordinate: $2 = \frac{3 + b + 1}{3}$ Multiplying both sides by 3: $6 = 3 + b + 1$ $6 = 4 + b \quad \Rightarrow \quad b = 2$

3. For the $z$-coordinate: $-1 = \frac{2 + (-4) + c}{3}$ Multiplying both sides by 3: $-3 = 2 - 4 + c$ $-3 = -2 + c \quad \Rightarrow \quad c = -1$

Thus, the values of $a$, $b$, and $c$ are: $a = 3, \quad b = 2, \quad c = -1$

Would you like more details on any of these steps?

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Here are 5 related questions you might find helpful:

1. How is the centroid formula derived from geometry?
2. What are the properties of a centroid in a triangle?
3. How does changing one vertex of the triangle affect the centroid's position?
4. Can the centroid ever lie outside the triangle?
5. How would the calculation change if the triangle was in 2D instead of 3D?

Tip: Always double-check your calculations for each coordinate when dealing with 3D geometry to avoid small errors.