Ans: D
Let $z = kx^3y^2$, where $k$ is a non-zero constant. Sub. $x=2$, $y=1$ and $z=14$ into $z=kx^3y^2$, we have
Let $z = kx^3y^2$, where $k$ is a non-zero constant. Sub. $x=2$, $y=1$ and $z=14$ into $z=kx^3y^2$, we have
$\begin{array}{rcl}
14 & = & k(2)^3(1)^2 \\
14 & = & 8k \\
k & = & \dfrac{7}{4} \\
\end{array}$
Therefore, $z = \dfrac{7}{4} x^3y^2$. For $x=3$ and $y=-2$,
$\begin{array}{rcl}
z & = & \dfrac{7}{4} \times (3)^3 \times (-2)^2 \\
z & = & 189
\end{array}$
