Finding minima from simultaneous equations We are given that a point $(x,y,z)$ in $\mathbb{R}^3$ satisfies the following equations
$x\cos\alpha-y\sin\alpha+z =1+\cos\beta$
$x\sin\alpha+y\cos\alpha+z =1-\sin\beta$
$x\cos(\alpha+\beta)-y\sin(\alpha+\beta)+z=2$
Where $\alpha,\beta\in\mathbb(0,2\pi)$
We need to find the Minimum value , $M$ of $x^2+y^2+z^2$.

\begin{align}M&=2\end{align}

My attempt:
By using Cramers rule, we find that the equations have an unique solution in the angle range mentioned. Solving for $x,y,z$ and then squaring and adding them and then finding the minimum value would take so much time. Is there any easier alternative way to do this?
 A: One can verify by inspection that
$$x=\cos(\alpha+\beta),\qquad y=-\sin(\alpha+\beta),\qquad z=1$$
solves the given system of equations, and when the system is regular there are no other solutions. This implies that $$x^2+y^2+z^2=2\ ,$$
independently of $\alpha$ and $\beta$; hence $M=2$.
But there is a catch: The determinant of the system computes to $1-\cos\beta+\sin\beta$, which is $=0$ for $\beta=0$ (excluded) and $\beta={3\pi\over2}\sim-{\pi\over2}$. When $\beta=-{\pi\over2}$ two of the equations coincide, so that we have an infinity of solutions, lying on a line in ${\mathbb R}^3$. Solving for $(x,y)$ in function of $z$ one obtains
$$x^2+y^2+z^2=5-6z+3z^2=2+3(z-1)^2\ ,$$
so that these special solutions do not falsify $M=2$.
A: When $\alpha = \beta = \pi$ and $x = 1, y = 0, z = 1$, the system of equations is met. So there exists $(x, y, z)$, which is a solution of the system of equations for some $\alpha, \beta \in (0, 2\pi)$, such that $x^2 + y^2 + z^2 = 2$.
On the other hand, by Cauchy-Bunyakovsky-Schwarz inequality, we have
\begin{align*}
&(x\cos (\alpha + \beta) - y\sin(\alpha + \beta) + z)^2 \\
\le\ & (x^2 + y^2 + z^2)(\cos^2 (\alpha + \beta) + \sin^2(\alpha + \beta) + 1^2 )\\ 
=\ & 2(x^2 + y^2 + z^2)
\end{align*}
which results in
$$x^2 + y^2 + z^2 \ge 2.$$
Thus, the minimum of $x^2 + y^2 + z^2$ is $2$.
