Since $U$ is unitary, $U^*=U^{-1}$. So, let us first think about how to find a matrix $U$ such that $U^{-1}AU$ is diagonal, without worrying about the condition that $U$ be unitary; we'll come back to fix that detail at the end.
Think about it this way: to diagonalize $A$, you want to find a change of basis: we know that the matrix for this transformation with respect to the basis $\{V_1,V_2\}$ is
$$
\begin{pmatrix}3 & 0\\0 & 0\end{pmatrix},
$$
because it maps $V_1$ to $3V_1$ and $V_2$ to $0V_2$.
If $U^{-1}AU$ is going to give you this, we want it to work like like so: when we plug in the vector $(1,0)^{T}$, we want $U$ to transform that in to $V_1$, so that $AU$ transforms it in to $3V_1$, so that $U^{-1}AU$ transforms it in to $(3,0)^T$. Does that make sense?
If $U$ is going to transform $(1,0)^T$ in to $V_1$, then the first column of $U$ must be exactly $V_1$. Similarly, the second column must be $V_2$. So, by this logic, we should take
$$
U=\begin{pmatrix}1+i & -1-i\\2 & 1\end{pmatrix}.
$$
For this matrix $U$, we certainly have $U^{-1}AU$ being the above matrix. However, there's one problem here: this matrix is not unitary.
So, instead of using $V_1$ and $V_2$ directly, let's use $W_1=\frac{V_1}{\|V_1\|}$ and $W_2=\frac{V_2}{\|V_2\|}$. If you take the matrix $U$ whose first column is $W_1$ and second column is $W_2$, you should get the properties you want.
In this case, we have
$$
\|V_1\|=\sqrt{(1+i)(1-i)+2^2}=\sqrt{6}\qquad \|V_2\|=\sqrt{(-1-i)(-1+i)+1^2}=\sqrt{3}
$$
and so we would take
$$
U=\begin{pmatrix}\frac{1+i}{\sqrt{6}} & \frac{-1-i}{\sqrt{3}}\\\frac{2}{\sqrt{6}} & \frac{1}{\sqrt{3}}\end{pmatrix}
$$
You can check that in this case, $U^*=U^{-1}$.
The searched matrix is $M=PDP^{-1}$ where $D$ is the diagonal matrix that has as diagonal elements the eigenvalues, in the same order as the eigenvectors in $P$ (see here).
$$
M=
\begin{bmatrix}
1&-1&0\\
0&1&0\\
2&0&1
\end{bmatrix}
\begin{bmatrix}
-1&0&0\\
0&0&0\\
0&0&1
\end{bmatrix}
\begin{bmatrix}
1&1&0\\
0&1&1\\
-2&-2&1
\end{bmatrix}
=
\begin{bmatrix}
-1&-1&0\\
0&0&0\\
-4&-4&1
\end{bmatrix}
$$
Best Answer
Now, you divide each of those vectors by its norm, thereby getting$$w_1=\frac13(-2i\ \ -2\ \ 1)^T,\ w_2=\frac13(2i\ \ -1\ \ 2)^T\text{, and }w_3=\frac13(-i\ \ 2\ \ 2)^T.$$And then you take$$C=\frac13\begin{bmatrix}-2i&2i&-i\\-2&-1&2\\1&2&2\end{bmatrix},$$whose columns are the vectors $w_1$, $w_2$, and $w_3$. And you're done.