= Solution
The angular-momentum <commutators> give
$$
\boxed{[L_+,L_z]=-\hbar L_+,\qquad
[L_-,L_z]=+\hbar L_-,\qquad
[L_+,L_-]=2\hbar L_z}.
$$
For example $[L_x+iL_y,L_z]=-i\hbar L_y+i(i\hbar L_x)=-\hbar L_+$. In the displayed basis $|1,1\rangle,|1,0\rangle,|1,-1\rangle$, the only nonzero upper entries of $L_+$ connect adjacent magnetic quantum numbers. Subtracting their column and row $L_z$ <eigenvalues> multiplies each by $-\hbar$, proving the first <matrix> <commutator>; the lower <matrix> similarly gives the second. Direct multiplication gives
$$
L_+L_-=2\hbar^2\operatorname{diag}(1,1,0),\quad
L_-L_+=2\hbar^2\operatorname{diag}(0,1,1),
$$
whose difference is $2\hbar L_z$. Also $L^2=L_z^2+(L_+L_-+L_-L_+)/2=2\hbar^2I$, confirming $\ell=1$.
A <rotation> changes the quantization axis without changing the spectrum, so \b[$L_{\mathbf n}$ has <eigenvalues> $+\hbar,0,-\hbar$]. Its dimensionless <matrix> is
$$
\frac{L_{\mathbf n}}{\hbar}=
\begin{pmatrix}
\cos\theta&\sin\theta/\sqrt2&0\\
\sin\theta/\sqrt2&0&\sin\theta/\sqrt2\\
0&\sin\theta/\sqrt2&-\cos\theta
\end{pmatrix}.
$$
Solving $(L_{\mathbf n}-\hbar I)v=0$ and normalizing gives the <spin-one maximum-projection eigenvector>
$$
\boxed{v_+(\theta)=
\begin{pmatrix}
\cos^2(\theta/2)\\
\sin\theta/\sqrt2\\
\sin^2(\theta/2)
\end{pmatrix}}.
$$
For instance the first equation is $(\cos\theta-1)\cos^2(\theta/2)+\sin^2\theta/2=0$, and the squared components sum to one. An overall <phase> is immaterial.
At $\theta=\pi/3$, $v_+=(3/4,\sqrt6/4,1/4)^{\mathsf T}$. The <Born rule> therefore predicts, in the requested order,
$$
\boxed{P(L_z=-\hbar)=\frac1{16},\qquad P(L_z=0)=\frac38,\qquad P(L_z=+\hbar)=\frac9{16}}.
$$
This treats the <Stern-Gerlach experiment> as the stipulated ideal measurement of orbital projection. In an actual atom, other <magnetic moments> and <spin-orbit coupling> can change which <angular momentum> the apparatus resolves.
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