The problem asks to evaluate the expression $(_{12}C_3) \cdot 7 \cdot (_ {12}C_3)$.

1. Problem Description

The problem asks to evaluate the expression $(_{12}C_3) \cdot 7 \cdot (_ {12}C_3)$.

2. Solution Steps

First, we need to calculate the combination $_{12}C_3$. The formula for combinations is:

$_nC_r = \frac{n!}{r!(n-r)!}$

Applying this formula to $_{12}C_3$, we get:

$_{12}C_3 = \frac{12!}{3!(12-3)!} = \frac{12!}{3!9!} = \frac{12 \times 11 \times 10 \times 9!}{3 \times 2 \times 1 \times 9!} = \frac{12 \times 11 \times 10}{6} = 2 \times 11 \times 10 = 220$

Now, substitute the value of $_{12}C_3$ into the expression:

$(_{12}C_3) \cdot 7 \cdot (_{12}C_3) = 220 \cdot 7 \cdot 220$

$= 220 \times 220 \times 7$

$= 48400 \times 7$

$= 338800$

3. Final Answer

338800