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*}
\begin{equation} \vec{F} \times \vec{R} = - \vec{M} \end{equation}
$\vec{R}$ 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.