If the expression $\frac{6 x^{2} y^{\frac{3}{2}}}{\sqrt[4]{81 x^{4} y}}$ is written in the form $a x^{b} y^{c}$, then what is the product of $a$, $b$ and $c$ ?

Step 1 ：Rewrite the expression inside the square root as \(81^{1/4} x^{4/4} y^{1/4}\), which simplifies to \(3 x y^{1/4}\).

Step 2 ：Substitute this into the original expression to get \(\frac{6 x^{2} y^{3/2}}{3 x y^{1/4}}\).

Step 3 ：Simplify this by subtracting the exponents of like terms in the numerator and denominator.

Step 4 ：The coefficient \(a\) is the ratio of the coefficients in the numerator and denominator, which is \(\frac{6}{3} = 2.0\).

Step 5 ：The exponent \(b\) of \(x\) is the difference of the exponents of \(x\) in the numerator and denominator, which is \(2 - 1 = 1\).

Step 6 ：The exponent \(c\) of \(y\) is the difference of the exponents of \(y\) in the numerator and denominator, which is \(1.5 - 0.25 = 1.25\).

Step 7 ：Finally, find the product of \(a\), \(b\), and \(c\), which is \(2.0 \times 1 \times 1.25 = 2.5\).

Step 8 ：Final Answer: The product of \(a\), \(b\), and \(c\) is \(\boxed{2.5}\).