$$\mid \sqrt{10 + 2\sqrt{6} + 2\sqrt{10} + 2\sqrt{15}} \mid + \mid \sqrt{10 - 2\sqrt{6} - 2\sqrt{10} + 2\sqrt{15}} \mid =$$

Firstly, we should know the identities:
$$\left(a+b+c\right)^2=a^2+b^2+c^2+2ab+2ac+2bc$$
$$\left(a-b-c\right)^2=a^2+b^2+c^2-2ab-2ac+2bc$$

In the question, the first expression inside the square root can be rewritten as:

$$10+2\sqrt{\ 6}+2\sqrt{\ 10}+2\sqrt{\ 15}=2+3+5+2\sqrt{\ 2}\sqrt{\ 3}+2\sqrt{\ 2}\sqrt{\ 5}+2\sqrt{\ 3}\sqrt{\ 5}$$=$$\left(\sqrt{\ 2}+\sqrt{\ 3}+\sqrt{\ 5}\right)^2$$
Hence, the first modulus will return the net value $$=\sqrt{\ 2}+\sqrt{\ 3}+\sqrt{\ 5}^{ }$$

Similarly, the expression in the second square root can be simplified as $$=\left(\sqrt{\ 2}-\sqrt{\ 3}-\sqrt{\ 5}\right)^2$$
As sqrt(3) and sqrt(5) are greater than sqrt(2), the modulus will change the sign of the expression finally and it will becomes $$=-\sqrt{\ 2}+\sqrt{\ 3}+\sqrt{\ 5}$$

Hence, the resultant value will be option A