Solved it: : Noticing that left-hand sides are symmetric in $a$, $b$, and $c$ (in that any relabelling of the variables in one of the left-hand sides gives one of the others), we multiply all three equations to get $(a^2b^2c^2)^2=(28)(27)(21)$, which implies $a^2b^2c^2=\sqrt{(28)(27)(21)}=\sqrt{(4\cdot 7)(9\cdot 3)(3\cdot 7.2\cdot 3\dot3\cdot 7=126.$ Dividing this equation by each of the three original equations gives $c^2 = 9/2$, $a^2 = 6$, and $b^2=14/3$, respectively. Since $b
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