In search for the maximum

Writing $\sqrt{3}a = 10\cos\alpha \hspace{1cm} \sqrt{3}b = 12\cos\beta \hspace{1cm} \sqrt{3}c = 14\cos\gamma$ the expression can be written as $20\sin(\alpha+\tfrac13\pi) + 24\sin(\beta + \tfrac13\pi) + 28\sin(\gamma + \tfrac13\pi)$ which clearly takes its maximum value of $\boxed{72}$ when $\alpha = \beta = \gamma = \tfrac16\pi$ , so when $(a,b,c) = (5,6,7)$ .

Splitting the terms with $a$ , $b$ , and $c$ separately and apply Cauchy-Schwarz inequality as follows:

$\begin{aligned} 3a + \sqrt{100-3a^2} & = a + a + a + \sqrt{100-3a^2} & \small \blue{\text{By Cauchy-Schwarz inequality}} \\ \left(a+a+a+\sqrt{100-3a^2} \right)^2 & \le 4 \left(a^2 + a^2 + a^2 + 100 - 3a^2 \right) = 400 \\ \implies 3a + \sqrt{100-3a^2} & \le 20 & \small \blue{\text{Equality occurs when }a = 5 } \end{aligned}$

Similarly, $3b + \sqrt{144-3b^2} \le 24$ and $3c + \sqrt{196-3c^2} \le 28$ and equalities occur when $b=6$ and $c=7$ respectively. Therefore,

$3(a+b+c) + \sqrt{100-3a^2} + \sqrt{144-3b^2} + \sqrt{196-3c^2} \le 20+24+28 = \boxed{72}$