Mathematicians usually are not worried about the conecpt of physical units. As such, a mathematician probably would argue that $\mathbf M_O$, $\vec{OP}$ and $\mathbf F$ belong to $\mathbb{R}^3$, as MyUserIsThis did in his comment.
If this is not satisfactory to you, you could consider three distinct fields of numbers, say $\mathbb{R}_F$ for forces, $\mathbb{R}_P$ for positions, and $\mathbb{R}_T$ for torques, and think of these fields as having the corresponding units associated with them.
Then, you can construct vector fields like $\mathbb{R}_P^D$ for $D$-dimensional positions from these fields, and endow them with the usual vector space operations, such as addition and scalar multiplication.
By definition, the vector spaces are not identical, but clearly they are isomorphic. To preserve the physical distinction between e.g. forces and torques, you would refrain from defining addition operations that take a pair $\left(\mathbb{R}_F^D,\mathbb{R}_T^D\right)$ to some other space. However, as a physicist, you would probably want to be able to form a cross product, i.e. a map $\left(\mathbb{R}_N^3,\mathbb{R}_P^3\right) \to \mathbb{R}_T^3$, just like you would desire to define a product $\left(\mathbb{R}_N,\mathbb{R}_P\right) \to \mathbb{R}_T$ that allows you to multiply scalar forces and distances to get something with the dimension of "space times force". (In fact, you probably would use the latter operation to define the former.)
This procedure has the advantage of being type-safe, to borrow a term from computer science, in the sense that you cannot add distances and forces "by accident", since there is no such addition operation defined. However, thinking along these lines may be tedious to some, and others may argue there is little mathematical reason to make the distinction between the different $\mathbb R$s, since they are isomorphic after all.
I hope you find these thoughts useful. I'd be happy to discuss them further if you so desire.