$ \vec u = (a, b, c) , \vec v = (b, a, 3) , \vec u . \vec v = {\vert u \vert}^2 , \vert \vec u - \vec v \vert = 5, c = ?$ $\vec u . \vec v = {\vert \vec u \vert}^2$
$\vert \vec u - \vec v \vert = 5$
Using these identities 
$\vec u . \vec v = \vert \vec u \vert . \vert \vec v \vert . \cos \theta$
${\vert \vec u + \vec v \vert }^2 - 2 \vert \vec u \vert . \vert \vec v \vert . \cos \theta = {\vert \vec u \vert}^2 + {\vert \vec v \vert}^2 $
without expanding $\vec u . \vec v  = ab + ab + 3c$ and $ \vec u - \vec v  = (a-b , b - a, c - 3)$ how do i continue?

$$\vec u = (a, b, c) , \vec v = (b, a, 3) , \vec u . \vec v = {\vert u \vert}^2 , \vert \vec u - \vec v \vert = 5, c = ?$$
$$\vec u . \vec v=ab+ba+3c$$
$${\vert u \vert}^2=a^2+b^2+c^2$$
$$|\vec u - \vec v |^2=(a-b)^2+(b-a)^2+(3-c)^2$$
We have $$2ab+3c=a^2+b^2+c^2$$
$$2(a-b)^2+(3-c)^2=25$$
or 
$$3c-c^2=(a-b)^2$$
$$2(a-b)^2+(3-c)^2=25$$
Subtract twice the first equation to the second.
$$(3-c)^2=25-2c(3-c)$$
or 
$$-16-c^2=0$$
you can conclude
