In a three-dimensional coordinate system, `P ,Q ,a n dR` are images of a point `A(a ,b ,c)` in the `x-y ,y-za n dz-x` planes, respectively. If `G` is the centroid of triangle `P Q R ,` then area of triangle `A O G` is (`O` is the origin) a. `0` b. `a^2+b^2+c^2` c. `2/3(a^2+b^2+c^2)` d. none of these

Options

A. 0
B. `a^(2)+b^(2)+c^(2)`
C. `2/3(a^(2)+b^(2)+c^(2))`
D. none of these

Explanation

To solve the problem, we need to find the area of triangle \( AOG \) where \( A \) is the point \( (a, b, c) \), \( O \) is the origin \( (0, 0, 0) \), and \( G \) is the centroid of triangle \( PQR \).

### Step 1: Determine the coordinates of points \( P, Q, R \)

1. **Point \( P \)** is the image of point \( A \) in the \( x-y \) plane. Therefore, the coordinates of \( P \) are:
\[ P = (a, b, 0) \]

2. **Point \( Q \)** is the image of point \( A \) in the \( y-z \) plane. Thus, the coordinates of \( Q \) are:
\[ Q = (0, b, c) \]

3. **Point \( R \)** is the image of point \( A \) in the \( z-x \) plane. Hence, the coordinates of \( R \) are:
\[ R = (a, 0, c) \]

### Step 2: Calculate the centroid \( G \) of triangle \( PQR \)

The coordinates of the centroid \( G \) of triangle \( PQR \) can be calculated using the formula:
\[ G = \left( \frac{x_1 + x_2 + x_3}{3}, \frac{y_1 + y_2 + y_3}{3}, \frac{z_1 + z_2 + z_3}{3} \right) \]
Substituting the coordinates of points \( P, Q, R \):
\[ G = \left( \frac{a + 0 + a}{3}, \frac{b + b + 0}{3}, \frac{0 + c + c}{3} \right) = \left( \frac{2a}{3}, \frac{2b}{3}, \frac{2c}{3} \right) \]

### Step 3: Determine the area of triangle \( AOG \)

The area of triangle formed by points \( A, O, G \) can be calculated using the formula for the area of a triangle in 3D:
\[ \text{Area} = \frac{1}{2} \| \vec{AO} \times \vec{AG} \| \]
Where:
- \( \vec{AO} = O - A = (0 - a, 0 - b, 0 - c) = (-a, -b, -c) \)
- \( \vec{AG} = G - A = \left( \frac{2a}{3} - a, \frac{2b}{3} - b, \frac{2c}{3} - c \right) = \left( -\frac{a}{3}, -\frac{b}{3}, -\frac{c}{3} \right) \)

### Step 4: Calculate the cross product \( \vec{AO} \times \vec{AG} \)

Calculating the cross product:
\[ \vec{AO} \times \vec{AG} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -a & -b & -c \\ -\frac{a}{3} & -\frac{b}{3} & -\frac{c}{3} \end{vmatrix} \]

Calculating this determinant:
\[ = \hat{i} \left( -b \cdot -\frac{c}{3} - -c \cdot -\frac{b}{3} \right) - \hat{j} \left( -a \cdot -\frac{c}{3} - -c \cdot -\frac{a}{3} \right) + \hat{k} \left( -a \cdot -\frac{b}{3} - -b \cdot -\frac{a}{3} \right) \]
All components will yield zero since they are equal and opposite, leading to:
\[ \vec{AO} \times \vec{AG} = (0, 0, 0) \]

### Step 5: Calculate the area

Since the cross product is zero, the area of triangle \( AOG \) is:
\[ \text{Area} = \frac{1}{2} \| \vec{AO} \times \vec{AG} \| = \frac{1}{2} \times 0 = 0 \]

### Conclusion

Thus, the area of triangle \( AOG \) is \( 0 \).

### Final Answer

The correct option is:
a. \( 0 \)