7. Given $x, y, z \in \mathbf{R}$, then $\sum \frac{x^{2}}{(3 x-2 y-z)^{2}}$ has the minimum value of $\qquad$ ("sum" indicates cyclic sum).


7. $\frac{5}{49}$.

Let $a=\frac{4 x+2 y+z}{7}, b=\frac{4 y+2 z+x}{7}$, $c=\frac{4 z+2 x+y}{7}$.
Then the original expression $=\frac{1}{49} \sum\left(\frac{2 a-b}{a-b}\right)^{2}$ $=\frac{1}{49}\left[5+\left(\sum \frac{a}{a-b}\right)^{2}\right] \geqslant \frac{5}{49}$.
Furthermore, when $x=-1, y=0, z=4$, the equality holds. Therefore, the minimum value is $\frac{5}{49}$.