First, simplify the expression \(7 \sqrt{12} \cdot 2 \sqrt{2}\).
Simplify \(\sqrt{12}\) and \(\sqrt{2}\):
\[ \sqrt{12} = \sqrt{4 \times 3} = \sqrt{4} \cdot \sqrt{3} = 2\sqrt{3} \]
\[ \sqrt{2} = \sqrt{2} \]
Substitute the simplified square roots back into the expression:
\[ 7 \cdot 2\sqrt{3} \cdot 2 \cdot \sqrt{2} = 7 \cdot 2 \cdot 2 \cdot \sqrt{3} \cdot \sqrt{2} \]
Combine the constants and the square roots:
\[ 7 \cdot 2 \cdot 2 = 28 \]
\[ \sqrt{3} \cdot \sqrt{2} = \sqrt{6} \]
Thus, the expression becomes:
\[ 28 \sqrt{6} \]
The simplified expression is:
\[ \boxed{28} \]
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