If `P` is any arbitrary point on the circumcirlce of the equllateral trangle of side length `l` units, then `| vec P A|^2+| vec P B|^2+| vec P C|^2` is always equal to `2l^2` b. `2sqrt(3)l^2` c. `l^2` d. `3l^2`

Options

A. `2l^(2)`
B. `2sqrt3l^(2)`
C. `l^(2)`
D. `3l^(2)`

Explanation

To solve the problem, we need to find the value of \( | \vec{PA} |^2 + | \vec{PB} |^2 + | \vec{PC} |^2 \) where \( P \) is any arbitrary point on the circumcircle of an equilateral triangle \( ABC \) with side length \( l \).

### Step-by-Step Solution:

1. **Understanding the Geometry**:
- Let \( A, B, C \) be the vertices of the equilateral triangle inscribed in a circumcircle with radius \( R \).
- The circumradius \( R \) of an equilateral triangle with side length \( l \) is given by the formula:
\[ R = \frac{l}{\sqrt{3}} \]

2. **Positioning the Triangle**:
- Place the triangle in the coordinate system such that the centroid (which is also the circumcenter for an equilateral triangle) is at the origin \((0, 0)\).
- The coordinates of the vertices can be represented as:
\[ A = \left( \frac{l}{2}, \frac{l\sqrt{3}}{6} \right), \quad B = \left( -\frac{l}{2}, \frac{l\sqrt{3}}{6} \right), \quad C = \left( 0, -\frac{l\sqrt{3}}{3} \right) \]

3. **Finding the Distances**:
- Let \( P \) be any point on the circumcircle. The squared distances from \( P \) to each vertex can be expressed as:
\[ | \vec{PA} |^2 = |P - A|^2, \quad | \vec{PB} |^2 = |P - B|^2, \quad | \vec{PC} |^2 = |P - C|^2 \]

4. **Using the Distance Formula**:
- The squared distance from point \( P(x, y) \) to point \( A \) is:
\[ | \vec{PA} |^2 = (x - \frac{l}{2})^2 + \left(y - \frac{l\sqrt{3}}{6}\right)^2 \]
- Similarly, calculate \( | \vec{PB} |^2 \) and \( | \vec{PC} |^2 \).

5. **Summing the Distances**:
- We need to sum these squared distances:
\[ | \vec{PA} |^2 + | \vec{PB} |^2 + | \vec{PC} |^2 \]

6. **Using the Property of the Circumcircle**:
- By the property of the circumcircle of an equilateral triangle, the sum of the squared distances from any point \( P \) on the circumcircle to the vertices \( A, B, C \) is constant and can be derived as:
\[ | \vec{PA} |^2 + | \vec{PB} |^2 + | \vec{PC} |^2 = 3R^2 + 3| \vec{OP} |^2 \]
- Since \( P \) lies on the circumcircle, \( | \vec{OP} | = R \).

7. **Substituting Values**:
- Substitute \( R = \frac{l}{\sqrt{3}} \):
\[ | \vec{PA} |^2 + | \vec{PB} |^2 + | \vec{PC} |^2 = 3 \left( \frac{l^2}{3} \right) + 3 \left( \frac{l^2}{3} \right) = 2l^2 \]

### Conclusion:
Thus, the expression \( | \vec{PA} |^2 + | \vec{PB} |^2 + | \vec{PC} |^2 \) is always equal to \( 2l^2 \).

### Final Answer:
The correct option is **a. \( 2l^2 \)**.