= Solution
The <operator commutator> is $[A,B]=AB-BA$, acting on a common domain where both products are defined. Use the <canonical commutation relation> $[x_i,p_j]=i\hbar\delta_{ij}$, with positions mutually commuting and momenta mutually commuting. On smooth test functions the only nonzero terms in the expansion are
$$
[yp_z,zp_x]=-i\hbar yp_x,\qquad [zp_y,xp_z]=i\hbar xp_y.
$$
Thus the <orbital angular momentum commutation relations> give
$$
\boxed{[L_x,L_y]=i\hbar(xp_y-yp_x)=i\hbar L_z,\qquad [L_i,L_j]=i\hbar\epsilon_{ijk}L_k.}
$$
The product rule for a <operator commutator> gives
$$
[L^2,L_w]=\sum_i\big(L_i[L_i,L_w]+[L_i,L_w]L_i\big)
=i\hbar\sum_{i,k}\epsilon_{iwk}(L_iL_k+L_kL_i)=0,
$$
because the last operator expression is symmetric in $i,k$, whereas $\epsilon_{iwk}$ is antisymmetric. Hence $\boxed{[L^2,L_w]=0}$ for every component.
Write $r=\sqrt{x^2+y^2+z^2}$ and $P=x+y+z$. The differential <angular momentum operator> $L=-i\hbar\,x\times\nabla$ annihilates every radial function, so $L_i(Pe^{-r})=e^{-r}L_iP$ and the same holds for its square. For example $L_xx=0$, $L_y^2x=\hbar^2x$ and $L_z^2x=\hbar^2x$. Cyclic symmetry gives $L^2y=2\hbar^2y$ and $L^2z=2\hbar^2z$ as well. Therefore
$$
\boxed{L^2\psi=e^{-r}L^2P=2\hbar^2\psi.}
$$
The wavefunction is an $l=1$ <orbital angular momentum> state; its nontrivial radial factor does not change its angular <eigenvalue>. The calculation holds away from the origin and defines the same angular-operator identity in the square-integrable state space.
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