If origin is the centroid of the triangle with vertices
P(3a, 3, 6), Q (-4, 2b, -8) and R(8, 12, 2c), then
the value of a, b, and c are

Options

A. `4/3, 1, 2`
B. `-4/3,-15/2,1`
C. `3, 2, -4/5`
D. `4/3, 15/2, 1`

Answer

To find the values of \( a \), \( b \), and \( c \) such that the origin is the centroid of the triangle with vertices \( P(3a, 3, 6) \), \( Q(-4, 2b, -8) \), and \( R(8, 12, 2c) \), we will use the formula for the centroid of a triangle in three-dimensional space.

### Step 1: Understanding the Centroid Formula
The centroid \( G \) 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:
\[ G = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}, \frac{z_1 + z_2 + z_3}{3} \right) \]
In our case, since the centroid is at the origin, \( G = (0, 0, 0) \).

### Step 2: Setting Up the Equations
From the centroid formula, we can set up the following equations for the coordinates:

1. For the x-coordinates:
\[ \frac{3a - 4 + 8}{3} = 0 \]
2. For the y-coordinates:
\[ \frac{3 + 2b + 12}{3} = 0 \]
3. For the z-coordinates:
\[ \frac{6 + (-8) + 2c}{3} = 0 \]

### Step 3: Solving the x-coordinate Equation
Starting with the x-coordinate equation:
\[ \frac{3a - 4 + 8}{3} = 0 \]
Multiply both sides by 3:
\[ 3a - 4 + 8 = 0 \]
Simplifying gives:
\[ 3a + 4 = 0 \implies 3a = -4 \implies a = -\frac{4}{3} \]

### Step 4: Solving the y-coordinate Equation
Next, we solve the y-coordinate equation:
\[ \frac{3 + 2b + 12}{3} = 0 \]
Multiply both sides by 3:
\[ 3 + 2b + 12 = 0 \]
Simplifying gives:
\[ 2b + 15 = 0 \implies 2b = -15 \implies b = -\frac{15}{2} \]

### Step 5: Solving the z-coordinate Equation
Finally, we solve the z-coordinate equation:
\[ \frac{6 - 8 + 2c}{3} = 0 \]
Multiply both sides by 3:
\[ 6 - 8 + 2c = 0 \]
Simplifying gives:
\[ -2 + 2c = 0 \implies 2c = 2 \implies c = 1 \]

### Final Values
Thus, the values of \( a \), \( b \), and \( c \) are:
\[ a = -\frac{4}{3}, \quad b = -\frac{15}{2}, \quad c = 1 \]