A particle of mass $m$ located at $\vec{r} = \pmatrix{x & y & z}$ from the origin, and having velocity $\vec{v} = \pmatrix{ vx & vy & vz} $ has the following properties
Linear momentum
$$\vec{p} = m \vec{v} = \pmatrix{m\, vx \\ m\, vy \\ m\, vz} $$
Angular momentum about the origin
$$ \vec{L} = \vec{r} \times \vec{p} = \pmatrix{m ( vz\, y-vy\, z) \\ m (vx\,z - vz\,x) \\ m( vy\,x -vx\,y )} $$
Where $\times$ is the vector cross product.
I think you are asking about finding the line of action of the reaction forces from an equipollent system of forces $\vec{F} = \pmatrix{Fx & Fy & Fz}$ and moments $\vec{M} = \pmatrix{Mx & My & Mz}$ at the location $\vec{r}$.
The is found with the calculation
$$ \vec{r}_{\rm zero moment} = \vec{r} + \frac{ \vec{F} \times \vec{M}} { \| \vec{F} \|^2} = \frac{1}{Fx^2+Fy^2+Fz^2} \pmatrix{Fy Mz-FzMy \\ Fz Mx - Fx Mz \\ Fx My - Fy Mx} $$
Anyway, I strongly suggest you do some reading on vectors and cross products in order to understand the math of mechanics.
My favorite way of calculating a cross product in via the cross-product matrix
$$ \vec{a} \times \vec{b} = \pmatrix{0 & -a_z & a_y\\ a_z & 0 & -a_x \\ -a_y & a_x & 0 } \pmatrix{b_x \\ b_y \\ b_z} $$
Use the shorthand notation $[\vec{a}\times]$ to denote the 3×3 skew symmetric matrix show above that multiplies $\vec{b}$. The above is concisely written as a matrix-vector multiplication which is easily computed
$$ \vec{a} \times \vec{b} = [\vec{a} \times ] \vec{b} $$
