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Employing the hint, the question mark should be replaced by $\sqrt{a^2+b^2-2ab\cos 60^{\circ}}=\sqrt{a^2+b^2-2ab\left(\dfrac{1}{2}\right)}=\sqrt{a^2+b^2-ab}$ on the basis of cosine rule. If $a$ were replaced by $c$ , we have $\sqrt{b^2+c^2-bc}$ for the other question mark. Now imagine $c$ were joined to the 60 degree vertex. Then $Q$ would be minimised if the two lengths with the question marks form a straight line when joined together. This minimum value would then be $\sqrt{a^2+c^2-2ab\cos 120^{\circ}}=\sqrt{a^2+c^2-2ac\left(-\dfrac{1}{2}\right)}=\sqrt{a^2+c^2+ac}$ .

So $q=\sqrt{a^2+c^2+ac}$ which means $q^2+ac=a^2+c^2+ac+ac=(a+c)^{2}=10^2=\boxed{100}$ .