Minimum Mean Square Estimator - Equivalent Expressions to Minimize
Since $ M \in \mathbb{S}^{N}_{++} $ (In other convention $ M \succ 0 $) by Cholesky Decomposition there is a Triangular Matrix $ R \in \mathbb{R}^{N \times N} $ such that $ M = {R}^{T} R $.

Using this fact one could prove $ 1 \iff 2 $ as following:

$$\begin{align*}
\arg \min_{\hat{x}} \mathbb{E} \left[ {\left( \hat{x} - x \right)}^{T} M \left( \hat{x} - x \right) \mid z \right] &amp; = \arg \min_{\hat{x}} \mathbb{E} \left[ {\left( \hat{x} - x \right)}^{T} {R}^{T} R \left( \hat{x} - x \right) \mid z \right] &amp;&amp; \text{$M = {R}^{T} R $} \\
&amp; = \arg \min_{\hat{x}} \mathbb{E} \left[ {\left( R \left( \hat{x} - x \right) \right)}^{T} \left( R \left( \hat{x} - x \right) \right) \mid z \right] &amp;&amp; \text{} \\
&amp; = \arg \min_{\hat{x}} \mathbb{E} \left[ {\left( R \hat{x} - y \right)}^{T} \left( R \hat{x} - y  \right) \mid z \right] &amp;&amp; \text{Defining $ y = R x $}
\end{align*}$$

Clearly, the above, as the classic MMSE problem, with respect to $ y $ given $ z $ which is minimized when:

$$ R \hat{x} = \mathbb{E} \left[ y \mid z \right] \Rightarrow \hat{x} = {R}^{-1} \mathbb{E} \left[ y \mid z \right] $$

Since $ M \succ 0 $ then $ {R}^{-1} $ is defined and the above is valid.
Moreover by linearity of the Expectation Operator:

$$ \hat{x} = {R}^{-1} \mathbb{E} \left[ y \mid z \right] = {R}^{-1} \mathbb{E} \left[ R x \mid z \right] = {R}^{-1} R \mathbb{E} \left[ x \mid z \right] = \mathbb{E} \left[ x \mid z \right] $$

Hence the minimizer of 1 indeed minimizes 2.

Using the cyclic property of the Trace Operator and the linearity of the Expectation Operator (Hence one could change the order of the $ \operatorname{Tr} \left( \cdot \right) $ and $ \mathbb{E} \left[ \cdot \right] $) one could easily show $ 2 \iff 3 $ hence the problem is solved.