The perimeter of the triangle above is 32 , so we get the length of $\(\mathrm{PQ}=32-(13+12)=32-25=7\)$

Now as $1\(3^2=169<12^2+7^2=144+49=193\)$ we get angle $\(Q\)$ less than 90 degrees.

[Note: - In a triangle say $\(\mathrm{ABC}, \mathrm{AC}^2=\mathrm{AB}^2+\mathrm{BC}^2 \Leftrightarrow \angle \mathrm{~B}=90^{\circ}, \mathrm{AC}^2>\mathrm{AB}^2+\mathrm{BC}^2 \Leftrightarrow \angle \mathrm{~B}>90^{\circ}\)$ \(&\) \(\mathrm{AC}^2<\mathrm{AB}^2+\mathrm{BC}^2 \Leftrightarrow \angle \mathrm{~B}<90^{\circ}\right]\)$

Hence column $\(B\)$ has higher quantity when compared with column $\(A\)$, so the answer is $\((B)\)$.