In all the $V_1$ ,$V_2$ ,$V_3$
$\bigg( \frac{7+8+15+23}{4} \bigg) ^2$=$\bigg( \frac{6+8+15+24}{4} \bigg) ^2$=$\bigg( \frac{5+8+15+25}{4} \bigg) ^2$=$53^{2}/4^{2}$
so it won't decide the relation.Hence no need to calculate it.
In all the $V_1$ ,$V_2$ ,$V_3$ ,first term is divided by 4 so it won't have impact on relation.Hence no need to calculate it.
also there is $8^{2}+15^{2}$ in numerator of first term of $V_1$ ,$V_2$ ,$V_3$. So it won't decide the relation.
Hence we need to calculate only $7^{2}+23^{2}$ in $V_1$, $6^{2}+24^{2}$ in $V_2$,$5^{2}+25^{2}$ in $V_3$
$7^{2}+23^{2}$=578,
$6^{2}+24^{2}$=612
$5^{2}+25^{2}$ =650
Therefore $V_3$ > $V_2$ >$V_1$