Solution:

We seek for integers $a, b, c$ such that $\sqrt{49+6 \sqrt{6}+12 \sqrt{14}+4 \sqrt{21}}=a \sqrt{2}+b \sqrt{3}+c \sqrt{7}$. The latter equation is equivalent to
$$\begin{aligned} 49+6 \sqrt{6}+12 \sqrt{14}+4 \sqrt{21} & =(a \sqrt{2}+b \sqrt{3}+c \sqrt{7})^{2} \\ & =2 a^{2}+3 b^{2}+7 c^{2}+2 a b \sqrt{6}+2 a c \sqrt{14}+2 b c \sqrt{21} \end{aligned}$$
Comparing similar terms, we see that $(a, b, c)$ is the solution to the following system of equations
$$2 a^{2}+3 b^{2}+7 c^{2} \stackrel{(1)}{=} 49, \quad 2 a b \stackrel{(2)}{=} 6, \quad 2 a c \stackrel{(3)}{=} 12, \quad 2 b c \stackrel{(4)}{=} 4$$
Multiplying equations (2), (3), (4) yields $8(a b c)^{2}=288$, so $a b c=6$. Hence, we get $c=2$ from (2), $b=1$ from (3) and $a=3$ from (4). Hence, $a+b+c=6$.