smc

If the value of $\frac{x^2+3xy}{y^2+z^2}$ is constant, as the answers imply, we can pick a value of $z$, and then solve the two linear equations for the corresponding $(x,y)$. We can then plug in $(x,y,z)$ into the expression to get the answer. If $z=1$, then $2x-3y=1$ and $x+3y=14$. We can solve each equation for $3y$ and set them equal, which leads to $2x-1=14-x$. This leads to $x=5$. Plugging in $x=5$ into $x+3y=14$ gives $y=3$. Thus, $(5,3,1)$ is one solution to the intersection of the two planes given. Plugging $(5,3,-1)$ into the expression gives $\frac{x^2+3xy}{y^2+z^2}$ gives $\frac{25+45}{10}$, or $7$, which is answer $\boxed{7}$