If $2x-3y-z=0$ and $x+3y-14z=0, z \neq 0$, the numerical value of $\frac{x^2+3xy}{y^2+z^2}$ is:
$\textbf{(A)}\ 7\qquad \textbf{(B)}\ 2\qquad \textbf{(C)}\ 0\qquad \textbf{(D)}\ -20/17\qquad \textbf{(E)}\ -2$
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 $\boxed{\textbf{(A)}}$