<s>[INST] 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?
 [/INST] 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 $2$.
</s>