The co-ordinates of a point `A_(n)` is `(n,n,sqrtn)` where `n in NN`, If `O(0,0,0)` is the origin then `sum_(i=1)^(12) (OA_(i))^(2)` equals `"_________".`

Answer

To solve the problem, we need to calculate the sum \( \sum_{i=1}^{12} (OA_i)^2 \) where the coordinates of point \( A_n \) are given as \( (n, n, \sqrt{n}) \) and \( O \) is the origin \( (0, 0, 0) \).

### Step-by-Step Solution:

1. **Identify the Coordinates**:
The coordinates of point \( A_n \) are \( (n, n, \sqrt{n}) \).

2. **Calculate the Distance \( OA_n \)**:
The distance \( OA_n \) from the origin \( O(0, 0, 0) \) to the point \( A_n(n, n, \sqrt{n}) \) can be calculated using the distance formula:
\[ OA_n = \sqrt{(n - 0)^2 + (n - 0)^2 + (\sqrt{n} - 0)^2} \]
Simplifying this gives:
\[ OA_n = \sqrt{n^2 + n^2 + (\sqrt{n})^2} = \sqrt{n^2 + n^2 + n} = \sqrt{2n^2 + n} \]

3. **Square the Distance**:
We need \( (OA_n)^2 \):
\[ (OA_n)^2 = 2n^2 + n \]

4. **Set Up the Summation**:
We need to find the sum:
\[ \sum_{i=1}^{12} (OA_i)^2 = \sum_{i=1}^{12} (2i^2 + i) \]
This can be separated into two sums:
\[ = \sum_{i=1}^{12} 2i^2 + \sum_{i=1}^{12} i \]

5. **Calculate the First Sum**:
The formula for the sum of the first \( n \) squares is:
\[ \sum_{i=1}^{n} i^2 = \frac{n(n+1)(2n+1)}{6} \]
For \( n = 12 \):
\[ \sum_{i=1}^{12} i^2 = \frac{12 \cdot 13 \cdot 25}{6} = 650 \]
Therefore:
\[ \sum_{i=1}^{12} 2i^2 = 2 \cdot 650 = 1300 \]

6. **Calculate the Second Sum**:
The formula for the sum of the first \( n \) natural numbers is:
\[ \sum_{i=1}^{n} i = \frac{n(n+1)}{2} \]
For \( n = 12 \):
\[ \sum_{i=1}^{12} i = \frac{12 \cdot 13}{2} = 78 \]

7. **Combine the Results**:
Now combine both sums:
\[ \sum_{i=1}^{12} (OA_i)^2 = 1300 + 78 = 1378 \]

### Final Answer:
Thus, the value of \( \sum_{i=1}^{12} (OA_i)^2 \) is \( 1378 \).