Oddly Positioned Table Legs

If the three thrusts are $T_1,T_2,T_3$ , the total moment on the table must be zero, and hence $\left(\begin{array}{c} 3 \\ 1 \\ 0 \end{array}\right) \wedge \left(\begin{array}{c} 0 \\ 0 \\ T_1\end{array}\right) \,+\, \left(\begin{array}{c} 5 \\ 2 \\ 0 \end{array}\right) \wedge \left(\begin{array}{c} 0 \\ 0 \\ T_2 \end{array}\right) \,+\, \left(\begin{array}{c} 1 \\ 5 \\ 0 \end{array}\right) \wedge \left(\begin{array}{c} 0 \\ 0 \\ T_3 \end{array}\right) \; = \; \left(\begin{array}{c} 3 \\ 3 \\ 0 \end{array}\right) \wedge \left(\begin{array}{c} 0 \\ 0 \\ 10 \end{array}\right)$ and the total force acting on the table is $0$ , so that $T_1 + T_2 + T_3 = 10$ . We obtain the simultaneous equations $\begin{aligned} T_1 + 2T_2 + 5T_3 & = \; 30 \\ 3T_1 + 5T_2 + T_3 & = \; 30 \\ T_1 + T_2 + T_3 & = \; 10 \end{aligned}$ and hence $T_1 = 2$ , $T_2 = T_3 = 4$ . Thus the answer is $\big\lfloor 100 \times \sqrt[3]{32}\big\rfloor = \boxed{317}$ .

Just for simplicity from the picture you see one of the leg is just on the center line of the table and the other two equidistant from it, which means that this two loads supported by these two legs are equals lets called R so 2R+Q =10, we have called Q the other load. Now taking moment to the horizontal axis of the table we get after simplifying R+2Q=0. So visually you get the answer R=4 and Q=2. Of course the very kind number for leg locations and size of table produces such easy solution. For general case the solution by Mark Henning is the path to analytically tackle the problem.