= Solution
The <matrix> sending the standard <basis> vectors to the given columns is
$$
\boxed{U=\frac1{\sqrt6}\begin{pmatrix}\sqrt3&-i\sqrt3&0\\1&i&-2\\1+i&-1+i&1+i\end{pmatrix}.}
$$
With the <Hermitian inner product>, the squared column <norms> are $3/6+1/6+2/6=1$, $3/6+1/6+2/6=1$, and $4/6+2/6=1$. The off-diagonal products are
$$
\begin{aligned}
\mathbf u_1^\dagger\mathbf u_2&=\frac{-3i+i+(1-i)(-1+i)}6=0,\\
\mathbf u_1^\dagger\mathbf u_3&=\frac{-2+(1-i)(1+i)}6=0,\\
\mathbf u_2^\dagger\mathbf u_3&=\frac{2i+(-1-i)(1+i)}6=0.
\end{aligned}
$$
Thus the columns form an <orthonormal basis>, so $\boxed{U^\dagger U=I}$ and $U$ is a <unitary matrix>. Here the operation maps the old <basis> vectors to the new vectors; the inverse coordinate <change of basis> uses $U^\dagger$.
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