In $\mathbb{R}^3$, we are asked to determine the matrix $P$, and its eigenvalues, corresponding to the projector of a vector on the plane orthogonal to $$\underline{v}=\left[\begin{matrix}1 \\ 2 \\ -3 \end{matrix}\right]$$
I think that the request is to determine the projector on the plane $W$ such that: $$W \oplus V = \mathbb{R^3}$$ where $V=\operatorname{span}(\underline{v})$ and $W=V^\perp$. A basis for $W$ could be: $$\{\underline{w}_1, \underline{w}_2\} = \{ \left[\begin{matrix} -2 \\ 1 \\ 0 \end{matrix}\right], \left[\begin{matrix} 3 \\ 0 \\ 1 \end{matrix}\right] \} $$ To extract a orthonormalized basis of $W$ I can apply Gram Schmidt algorithm. So a vector projected on $W$: $$x_w= (u_1, x)u_1+(u_2, x)u_2$$ Where $\{u_1, u_2\}$ are orthonormalized vectors basis of $W$. How can I find the matrix of the projection?
After some pther researches I found that $$P= u_1 u_1^T + u_2u_2^T$$ So I ask: if this proceeding is correct, is there an easier way to get this result?
You have given two methods, both correct. For the first method, you have correctly expressed your projection $P$ as a linear operator. To find the matrix with respect to the standard basis, use the rule that the $j$-th column is $P(e_j)$ where $e_j$ is the $j$-th standard basis element. You can check, both conceptually and computationally that this agrees with your second formula.
Yet another method, which avoids the computation of $u_1, u_2$ entirely, so is probably better: the projector onto the plane perpendicular to $v$ is the identity minus the projector onto the span of $v$, $$ P(x) = x - \frac{\langle x, v \rangle }{||v||^2} v. $$ Again to find the matrix, the same rule applies. The the $j$-th column is $P(e_j)$. Again, you can check by computation that this will agree with your other methods.