Let G be the centroid of the `DeltaABC`, whose sides are of lengths a,b,c. If P be a point in the plane of `triangleABC`, such that `PA=1,PB=3, PC=4` and `PG=2`, then the value of `a^(2)+b^(2)+c^(2)` is

Options

A. 42
B. 40
C. 36
D. 28

Explanation

To solve the problem, we will use the properties of the centroid and the distances from point P to the vertices of triangle ABC.

### Step-by-Step Solution:

1. **Understanding the Given Data**:
We have a triangle ABC with centroid G. The distances from point P to the vertices are given as:
- \( PA = 1 \)
- \( PB = 3 \)
- \( PC = 4 \)
- \( PG = 2 \)

2. **Using the Formula for Distances**:
We can use the formula that relates the distances from an arbitrary point to the vertices of a triangle and the centroid:
\[ PA^2 + PB^2 + PC^2 = 3PG^2 + \frac{1}{3}(a^2 + b^2 + c^2) \]

3. **Substituting the Known Values**:
We will substitute the known values into the formula:
\[ PA^2 = 1^2 = 1, \quad PB^2 = 3^2 = 9, \quad PC^2 = 4^2 = 16 \]
Thus,
\[ PA^2 + PB^2 + PC^2 = 1 + 9 + 16 = 26 \]

4. **Substituting PG**:
Now substitute \( PG = 2 \):
\[ PG^2 = 2^2 = 4 \]
Therefore,
\[ 3PG^2 = 3 \times 4 = 12 \]

5. **Setting Up the Equation**:
Now we can set up the equation:
\[ 26 = 12 + \frac{1}{3}(a^2 + b^2 + c^2) \]

6. **Solving for \( a^2 + b^2 + c^2 \)**:
Rearranging the equation gives:
\[ 26 - 12 = \frac{1}{3}(a^2 + b^2 + c^2) \]
\[ 14 = \frac{1}{3}(a^2 + b^2 + c^2) \]
Multiplying both sides by 3:
\[ a^2 + b^2 + c^2 = 42 \]

7. **Final Answer**:
Therefore, the value of \( a^2 + b^2 + c^2 \) is:
\[ \boxed{42} \]