Solution of Vector Cross Product of Different Vectors

Although the solution of a cross-product identity is crystal clear here, my question differs from the asked ones. Let's assume that I have three force vectors and these vectors are applied at different locations. They are given as:

\begin{align*} \vec{F_1} = \pmatrix{5 \\ 22 \\ 9}, \vec{F_2} = \pmatrix{8 \\ 5 \\ 11}, \vec{F_3} = \pmatrix{9 \\ 21 \\ 7} \end{align*}

\begin{align*} \vec{r_1} = \pmatrix{1 \\ 0.75 \\ -6.23}, \vec{r_2} = \pmatrix{7.12 \\ 5.63 \\ 8.45}, \vec{r_3} = \pmatrix{-9.45 \\ -1.12 \\ 6.11} \end{align*}

I can find the equivalent force and moment at the origin $$(0,0,0)$$ as follows:

\begin{align*} \vec{F} = \vec{F_1} + \vec{F_2} + \vec{F_3} = \pmatrix{ 22 \\ 48 \\ 27} \end{align*}

\begin{align*} \vec{M} = \vec{r_1} \times \vec{F_1} + \vec{r_2} \times \vec{F_2} + \vec{r_3} \times \vec{F_3} = \pmatrix{ 27.34 \\ 70.27 \\ -179.56} \end{align*}

From now on, I would like to find a point (which are infinitely many) to yield zero moment. Moment is a free vector but force is not. Denoting such point at $$(x_0,y_0,z_0)$$, I need to solve the following equation:

\begin{align*} -\vec{R} \times \vec{F} + \vec{M} = \vec{0} \end{align*}

$$$$\vec{F} \times \vec{R} = - \vec{M}$$$$

$$\vec{R}$$ being my unknown, this is a vector equation whose solution is actually given as:

\begin{align*} \vec{R} = \frac{-\vec{M} \times \vec{F}}{\vec{F} \cdot \vec{F}} + k \vec{F} \end{align*}

where $$k$$ is an arbitrary scalar and hence, $$\vec{R}$$ is infinitely many. However, if you check, $$\vec{F}$$ is not normal to $$\vec{M}$$ and for these values I cannot solve the equation because the cross product identity is not satisfied. How come the cross product equation indicates that this is vice versa? What I am missing here?

The cross-product is orthogonal to the two vectors in the product. So for a vector $$\vec R$$ to exist for which $$\vec M = \vec R \times \vec F$$, $$\vec M$$ has to be orthogonal to $$\vec F$$.
When $$\vec F$$ and $$\vec M$$ are orthogonal, then $$\vec F$$ is on the plane normal to $$\vec M$$. For any other vector on that plane not parallel to $$\vec F$$, the cross-product with $$\vec F$$ will be parallel to $$\vec M$$, and some scalar multiple will give $$\vec M$$ itself. Thus there are infinitely many solutions for $$\vec R$$.
But when $$\vec F$$ and $$\vec M$$ are not orthogonal, no such $$\vec R$$ can exist. There is just no vector that will solve the equation.
Your solution works fine when $$\vec F$$ is orthogonal to $$\vec M$$. Because then $$\vec F \times (\vec M\times \vec F) = \|\vec F\|^2\vec M$$. But if $$\vec F$$ is not orthogonal to $$\vec M$$, it is not a multiple of $$\vec M$$ at all.