Clearly: U contains only vectors from V that have z-coordinate 0, thus if you take any linear combination $\lambda (1,2,3)+\mu(i,-i,10)$ from V, you have to require that $\lambda \cdot 3 +\mu\cdot 10=0$, hence $\lambda=-\frac{10}{3}\mu$.
This means that $U\cap V$ is 1-dimensional and that any single vector that you get by chosing $\mu,\lambda \neq 0$ and $\lambda=-\frac{10}{3}\mu$ will therefore span that intersection.
Take for example, $\mu=3$, hence $\lambda=-\frac{10}{3}\cdot 3 = -10$. This gives the vector
$-10(1,2,3)+3(i,-i,10)=(-10+3i,-20-3i,0)$ of $U\cap V$.