If the centroid of a triangle whose vertices are (a, 1, 3), (-2, b, -5) and (4, 7, c) is the origin, then the values of a, b, c are
A. (-2, -8, -2)
B. (2, 8, -2)
C. (-2, -8, 2)
D. (7, -1, 0)

Hint: The centroid of a triangle whose vertices are given by the coordinates $({x_1},{y_1},{z_1}),\,({x_2},{y_2},{z_2})\,{\text{and}}\,({x_3},{y_3},{z_3})$ is given by the formula, $G = (\dfrac{{{x_1} + {x_2} + {x_3}}}{3},\dfrac{{{y_1} + {y_2} + {y_3}}}{3},\dfrac{{{z_1} + {z_2} + {z_3}}}{3})$. Solve the three linear equations in one variable to get the values of a, b and c.

Formula Used:
$G = (\dfrac{{{x_1} + {x_2} + {x_3}}}{3},\dfrac{{{y_1} + {y_2} + {y_3}}}{3},\dfrac{{{z_1} + {z_2} + {z_3}}}{3})$.

Complete step by step solution:
We know that the centroid of a triangle whose vertices are given by the coordinates $({x_1},{y_1},{z_1}),\,({x_2},{y_2},{z_2})\,{\text{and}}\,({x_3},{y_3},{z_3})$ is given by the formula, $G = (\dfrac{{{x_1} + {x_2} + {x_3}}}{3},\dfrac{{{y_1} + {y_2} + {y_3}}}{3},\dfrac{{{z_1} + {z_2} + {z_3}}}{3})$
$(0,0,0) = (\dfrac{{a - 2 + 4}}{3},\dfrac{{1 + b + 7}}{3},\dfrac{{3 - 5 + c}}{3})$
$(0,0,0) = (\dfrac{{2 + a}}{3},\dfrac{{8 + b}}{3},\dfrac{{ - 2 + c}}{3})$
$\dfrac{{2 + a}}{3} = 0 \Rightarrow 2 + a = 0 \Rightarrow a = - 2$
$\dfrac{{8 + b}}{3} = 0 \Rightarrow 8 + b = 0 \Rightarrow b = - 8$
$\dfrac{{ - 2 + c}}{3} = 0 \Rightarrow - 2 + c = 0 \Rightarrow c = 2$
Therefore, a = -2, b = -8, c = 2

The correct option is option (C) (-2, -8, 2)

Note: The centroid of a triangle is a point inside the triangle from which all the vertices are equidistant. If any of the coordinates of the centroid of the triangle does not lie between the smallest and largest values of that coordinate in the three vertices, then there must be a calculation error.