Answer
$2x^{2}y$
Work Step by Step
Based on the quotient rule for exponents, we know that $\frac{a^{m}}{{a}^{n}}=a^{m-n}, a\ne0$ (where $m$ and $n$ are positive integers and $a$ is a real number).
Therefore, $\frac{2x^{3}y^{2}z}{xyz}=2\times x^{3-1}\times y^{2-1}\times z^{1-1}=2\times x^{2}\times y^{1}\times z^{0}=2x^{2}y$.
Recall that $z^{0}=1$.