Question

If the centroid of triangle whose vertices are \[ (a,1,3),\ (-2,b,-5)\ \text{and}\ (4,7,c) \] be the origin, then \[ a^2+b^2+c^2= \] is:

• A. \(68\)
• B. \(64\)
• C. \(72\)
• D. \(54\)

Hint:

The centroid of a triangle in three dimensions is obtained by taking the average of corresponding coordinates of the vertices.

Answer 1

The Correct Option is C

Step 1: Recall the centroid formula in 3D.
The centroid of a triangle with vertices $(x_1,y_1,z_1)$, $(x_2,y_2,z_2)$, $(x_3,y_3,z_3)$ is $\left(\dfrac{x_1+x_2+x_3}{3}, \dfrac{y_1+y_2+y_3}{3}, \dfrac{z_1+z_2+z_3}{3}\right)$.

Step 2: Set up equations for the x-coordinates.
Centroid x-coordinate $= 0$: $\dfrac{a + (-2) + 4}{3} = 0 \Rightarrow a + 2 = 0 \Rightarrow a = -2$.

Step 3: Set up equations for the y-coordinates.
$\dfrac{1 + b + 7}{3} = 0 \Rightarrow 8 + b = 0 \Rightarrow b = -8$.

Step 4: Set up equations for the z-coordinates.
$\dfrac{3 + (-5) + c}{3} = 0 \Rightarrow -2 + c = 0 \Rightarrow c = 2$.

Step 5: Compute $a^2 + b^2 + c^2$.
\[ a^2 + b^2 + c^2 = (-2)^2 + (-8)^2 + 2^2 = 4 + 64 + 4 = 72. \]

Step 6: State the answer.
$a^2 + b^2 + c^2 = 72$.
\[ \boxed{72} \]