= Solution
Use the radial exponential $e^{-\alpha r}$ shown in the original PDF; the TeX's $e^{-\alpha x}$ is a transcription defect. Every orbital <angular momentum operator> annihilates a radial function, so it acts only on $P=x_1+x_2+2x_3$. The <angular momentum of a radial function times a linear polynomial> is determined by its degree-one angular factor: direct application of the rotation derivatives gives
$$
\hat L^2P=2\hbar^2P.
$$
Equivalently its restriction to the sphere is a degree-one <spherical harmonic>. Thus
$$
\boxed{\hat L^2\psi=2\hbar^2\psi,\qquad l(l+1)=2,\quad l=1.}
$$
Furthermore $\hat L_3\psi=-i\hbar K(x_1-x_2)e^{-\alpha r}$. Its <expected value> is proportional to the integral of $(x_1+x_2+2x_3)(x_1-x_2)e^{-2\alpha r}$. The cross terms are odd in a coordinate, and the integrals of $x_1^2$ and $x_2^2$ agree by rotational symmetry. Hence
$$
\boxed{\langle\hat L_3\rangle_\psi=0.}
$$
The state need not be an eigenstate of $\hat L_3$. If normalization is desired, $\int P^2e^{-2\alpha r}\,d^3x=6\pi/\alpha^5$, so $K=\alpha^{5/2}/\sqrt{6\pi}$; the zero expectation holds after normalization for any nonzero initial $K$.
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