To find the product of the expression \(\sqrt{212} \cdot \sqrt{27} \cdot \sqrt{225}\), we can first combine the square roots:
\[ \sqrt{212} \cdot \sqrt{27} \cdot \sqrt{225} = \sqrt{212 \cdot 27 \cdot 225} \]
Now, we will calculate each component:
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\( \sqrt{212} = \sqrt{4 \cdot 53} = 2\sqrt{53} \)
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\( \sqrt{27} = \sqrt{9 \cdot 3} = 3\sqrt{3} \)
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\( \sqrt{225} = \sqrt{15^2} = 15 \)
Now we can substitute these back into our product:
\[ \sqrt{212} \cdot \sqrt{27} \cdot \sqrt{225} = (2\sqrt{53})(3\sqrt{3})(15) \]
Now, multiply the coefficients:
\[ 2 \cdot 3 \cdot 15 = 90 \]
Now multiply the square roots:
\[ \sqrt{53} \cdot \sqrt{3} = \sqrt{159} \]
Therefore, the complete expression becomes:
\[ 90\sqrt{159} \]
To put this in simplest radical form, we notice that \(90\) can be factored as \(9 \cdot 10\) but cannot simplify further with \(\sqrt{159}\) since \(159\) has a prime factorization of \(3 \times 53\).
Thus, the final answer is:
\[ 90\sqrt{159} \]
None of the provided responses seem to match this calculation, suggesting there may be an issue with the options given. However, based on the computations, the simplest radical form of the product is \(90\sqrt{159}\).