Chapter 6 - Section 6.7 - Variation and Problem Solving - Exercise Set - Page 398: 38

$k=10$

Since $y$ varies jointly as $x$ and the square of $z$, then, $ y=kxz^2 .$ Substituting the given values of the variables, then the value of $k$ is, \begin{array}{l} 360=k(4)(3)^2 \\\\ 360=k(4)(9) \\\\ 360=k(36) \\\\ \dfrac{360}{36}=k \\\\ 10=k \\\\ k=10 .\end{array}