The problem is to simplify the expression $(2x^{-1}y^{-2}z^{-3})^{-2} \times (3x^{-2}y^{-3}z^{-1})^3$ and write the answer with positive exponents.

1. Problem Description

The problem is to simplify the expression $(2x^{-1}y^{-2}z^{-3})^{-2} \times (3x^{-2}y^{-3}z^{-1})^3$ and write the answer with positive exponents.

2. Solution Steps

First, we apply the power of a product rule: $(ab)^n = a^n b^n$.

$(2x^{-1}y^{-2}z^{-3})^{-2} = 2^{-2}(x^{-1})^{-2}(y^{-2})^{-2}(z^{-3})^{-2} = 2^{-2}x^{2}y^{4}z^{6}$

$(3x^{-2}y^{-3}z^{-1})^3 = 3^3(x^{-2})^3(y^{-3})^3(z^{-1})^3 = 3^3x^{-6}y^{-9}z^{-3}$

Next, we multiply the two simplified expressions:

$(2^{-2}x^{2}y^{4}z^{6})(3^3x^{-6}y^{-9}z^{-3}) = 2^{-2}3^3x^{2-6}y^{4-9}z^{6-3} = 2^{-2}3^3x^{-4}y^{-5}z^{3}$

Now, we write the expression with positive exponents. Recall that $a^{-n} = \frac{1}{a^n}$.

$2^{-2}3^3x^{-4}y^{-5}z^{3} = \frac{3^3z^3}{2^2x^4y^5} = \frac{27z^3}{4x^4y^5}$

3. Final Answer

$\frac{27z^3}{4x^4y^5}$