Length of vector $p=a+b+c$ when $a,b,c$ lengths are provided and $a,b,c$ make $60^\circ$ angles with each other

The problem: Vectors $$a$$, $$b$$, $$c$$ make $$60^\circ$$ angles with each other. $$|a| = 4$$, $$|b| = 2$$, $$|c| = 6$$. Find the length of $$p = a + b + c$$.

The only way I can think of $$a$$, $$b$$ and $$c$$ having $$60^\circ$$ angles with each other is that they form a vertex of a tetrahedron. Then, I can find $$|a+b|$$ or $$|b+c|$$ or $$|a+c|$$ using the law of cosines. But then I can't find $$|p|$$, because I don't know the angle between the vector I have found and the remaining one.

I would like to get some hints or clues how to solve this, thanks in advance.

I would square the given sum: $$\vec{p}^2=(\vec{a}+\vec{b}+\vec{c})^2=\vec{a}^2+\vec{b}^2+\vec{c}^2+2\vec{a}\cdot \vec{b}+2\vec{b}\cdot \vec{c}+2\vec{c}\cdot\vec{a}$$ This is equal $$|\vec{a}|^2+|\vec{b}|^2+|\vec{c}|^2+2|\vec{a}||\vec{b}|\cos(\pi/3)+2|\vec{b}||\vec{c}|\cos(\pi/3)+2|\vec{a}||\vec{c}|\cos(\pi/3)$$