We assume the beginning of the coordinate system at point A.
We will calculate the moment by two methods - with respect to point A by the first method, and with respect to point B by the second method.
Reduction at point A (Method I). In vector notation, the moment of force with respect to a point is defined as

\begin{aligned} &\overline{M_A}=\sum_{i=1}^n{(\overline{r_i}\times\overline{P_i})} \end{aligned}
where $r_i$ is the radius vector - the displacement vector from the point of force attachment to the moment pole.
So we number the forces, denote the attachment points of these forces
\begin{aligned} &P_1=4\sqrt{2} \ kN; &&A_1=(-2,-2,0) m\\ &P_2=3\sqrt{13} \ kN ; &&A_2=(3,0,0) m\\ &P_3=\sqrt{5} \ kN; &&A_3=(4,0,0) m\\ &P_4=2 \ kN; &&A_4=(4,-4,0) m\\ &P_5=2\sqrt{13} \ kN ; &&A_5=(3,-4,0) m\\ &P_6=10 \ kN ; &&A_6=(-2,-4,0) m\\ \\ &\text{Coordinates of point A}\\ &A=(0,0,0)m\\ \\ &\text{Radius vectors}\\ &\overline{A_1A}=(2,2,0) m\\ &\overline{A_2A}=(-3,0,0) m\\ &\overline{A_3A}=(-4,0,0) m\\ &\overline{A_4A}=(-4,4,0) m\\ &\overline{A_5A}=(-3,4,0) m\\ &\overline{A_6A}=(2,4,0) m\\ \end{aligned}
We need to project the force onto the x and y directions, this can also be done in at least two ways, we will show the first method with Method I and the second method with Method II.
\begin{aligned} &\overline{P_i}=|P_i|\cdot\frac{\overline{r_i}}{|r_i|}\\ &\overline{r_i} - \text{a vector specifying the direction of the force } \overline{P_i}\\ \\ &\overline{P_1}=|P_1|\cdot\frac{\overline{A_1A}}{|A_1A|}\\ &\overline{P_1}=4\sqrt{2}\cdot\frac{(2,2,0)}{\sqrt{2^2+2^2+0^2}}=\frac{4\sqrt{2}(2,2,0)}{2\sqrt{2}}=(4,4,0) \ kN\\ &\overline{P_2}=|P_2|\cdot\frac{\overline{A_2C}}{|A_2C|}\\ &\overline{P_2}=3\sqrt{13}\cdot\frac{(-3,-2,0)}{\sqrt{9+4}}=(-9,-6,0) \ kN\\ &\overline{P_3}=|P_3|\cdot\frac{\overline{A_3B}}{|A_3B|}\\ &\overline{P_3}=\sqrt{5}\cdot\frac{(-1,-2,0)}{\sqrt{1+4}}=(-1,-2,0) \ kN\\ &\overline{P_4}=(0,2,0) \ kN\\ &\overline{P_5}=|P_5|\cdot\frac{\overline{A_5C}}{|A_5C|}\\ &\overline{P_5}=2\sqrt{13}\cdot\frac{(-3,2,0)}{\sqrt{9+4}}=(-6,4,0) \ kN\\ &\overline{P_6}=(10,0,0) \ kN\\ \end{aligned} \begin{aligned} &\text{Moment with respect to point A}\\ &\overline{M_1}=\overline{P_1}\times \overline{A_1A}=&(4,4,0)\\ &&\underline{(2,2,0)}\\ &&(0,0,0)\\ \\ &\overline{M_2}=\overline{P_2}\times \overline{A_2A}=&(-9,-6,0)\\ &&\underline{(-3,0,0)}\\ &&(0,0,-18) \\ \\ &\overline{M_3}=\overline{P_3}\times \overline{A_3A}=&(-1,-2,0)\\ &&\underline{(-4,0,0)}\\ &&(0,0,-8) \\ \\ &\overline{M_4}=\overline{P_4}\times \overline{A_4A}=&(0,2,0)\\ &&\underline{(-4,4,0)}\\ &&(0,0,8) \\ \\ &\overline{M_5}=\overline{P_5}\times \overline{A_5A}=&(-6,4,0)\\ &&\underline{(-3,4,0)}\\ &&(0,0,-12) \\ \\ &\overline{M_6}=\overline{P_6}\times \overline{A_6A}=&(10,0,0)\\ &&\underline{(2,4,0)}\\ &&(0,0,40) \\ \\ &\overline{M_7}=&(0,0,-16)\\ &\overline{M_8}=&(0,0,2)\\ &\overline{M_A}=\sum_{i=1}^n{M_i}\\ &\overline{M_A}=(0,0,-4) \ kNm\\ \end{aligned}
Resultant force vector
\begin{aligned} &\overline{S}=\sum_{i=1}^n{\overline{P_i}}\\ &\overline{S}=&(4,4,0)\\ &&(-9,-6,0)\\ &&(-1,-2,0)\\ &&(0,2,0)\\ &&(-6,4,0)\\ &&\underline{(10,0,0)}\\ &\overline{S}=&(-2,2,0) & \ kN\\ \end{aligned}
The system at point A is reduced to the resultant $\overline{S}=(-2,2,0) \ kN$ and moment $\overline{M_A}=(0,0,-4)$

Reduction at point B (Method II)
Expand

One of the most popular and simplest methods of calculating a moment is to decompose forces into components and determine the moment each component makes with respect to a given point. The x-components make a moment on the y-axis and vice versa. All this is done by looking at the dimensions of the system, the values and directions of the forces.
To do this, we decompose forces into components, we will do it by the second method, also more common in subsequent sections of mechanics and strength of materials, we describe the triangle on which the force acts and we project the force onto the x and y axes. The force directions are marked on the figure 4, the values are calculated below.
$\begin{aligned} &\text{Force P1 is based on a right triangle 2×2, so it is inclined at an angle of 45 degrees}\\ &sin(45)=cos(45)=\frac{\sqrt{2}}{2}\\ &P_{1x}=P_{1y}=4\sqrt{2}\cdot\frac{\sqrt{2}}{2}=4 \ kN\\ \\ &\text{Angle } \beta:\\ &sin(\beta)=\frac{2}{\sqrt{2^2+3^2}}=\frac{2}{\sqrt{13}}\\ &cos(\beta)=\frac{3}{\sqrt{2^2+3^2}}=\frac{3}{\sqrt{13}}\\ &\text{Angle } \gamma:\\ &sin(\gamma)=\frac{2}{\sqrt{2^2+1^2}}=\frac{2}{\sqrt{5}}\\ &cos(\gamma)=\frac{1}{\sqrt{2^2+1^2}}=\frac{1}{\sqrt{5}}\\ \\ &P_{2x}=P_2\cdot cos(\beta)=9 \ kN\\ &P_{2y}=P_2\cdot sin(\beta)=6 \ kN\\ \\ &P_{3x}=P_3\cdot cos(\gamma)=1 \ kN\\ &P_{3y}=P_3\cdot sin(\gamma)=2 \ kN\\ \\ &P_{4y}=P_4=2 \ kN\\ \\ &P_{5x}=P_5\cdot cos(\beta)=6 \ kN\\ &P_{5y}=P_5\cdot sin(\beta)=4 \ kN\\ \\ &P_{6x}=P_6=10 \ kN\\ \end{aligned}$ $\begin{aligned} &M_{Bz}=-P_{1y}\cdot 5+P_{2x}\cdot 2 + P_{3x}\cdot 2-P_{3y}\cdot 1+P_4\cdot 1-P_{5x}\cdot 2+P_6 \cdot 2-M_7+M_8\\ &M_{Bz}=-4\cdot 5+9\cdot 2+1\cdot 2-2\cdot 1+2\cdot 1-6\cdot 2+10\cdot 2-16+2=-6 \ kNm\\ \end{aligned}$
The system at point B is reduced to the resultant $\overline{S}=(-2,2,0) \ kN$ and moment $\overline{M_B}=(0,0,-6)$

Check - theorem of changing the pole of the moment
Expand

The moment at point B can be calculated by the second method, with the theorem of changing the pole of the moment and thus verify the correctness of the calculations.
$\begin{aligned} &\overline{M_B}=\overline{M_A}+\overline{S}\times\overline{AB}=&(0,0,-4)+&(-2,2,0)\\ &&&\underline{(3,-2,0)}\\ &&&(0,0,-2)\\ &\overline{M_B}=(0,0,-4)+(0,0,-2)=(0,0,-6)\ kNm\\ &\text{the check gives a positive result}\\ \end{aligned}$

Reduction to the simplest form
Expand

As we know from the theoretical introduction, if $\overline{S}\ne 0, \overline{M_A}\ne 0$ then the system is reduced to the resultant, composed of a single non-zero vector equal to the sum of $\overline{S}$ lying on the central axis with the desired equation.
We determine the equation of the central axis
From the theorem of pole change
$\begin{aligned} &\overline{M_O}=\overline{M_A}+\overline{S}\times\overline{AO}\\ &\text{where: } &M_O- \text{zero moment}\\ &&O=(x,y,0)- \text{point with unknown coordinates}\\ \\ &\overline{AO}=(x,y,0) \\ \end{aligned}$ $\begin{aligned} &\overline{M_O}=\overline{M_A}+\overline{S}\times\overline{AO}=&(0,0,-4)+&(-2,2,0)\\ &&&\underline{(x,y,0)}\\ &&&(0,0,-2y-2x)\\ \\ &(0,0,0)=(0,0,-4)+(0,0,-2y-2x)\\ \end{aligned}$ $\begin{aligned} &\text{What happens if the equations are satisfied:}\\ &\begin{cases} &0=0+0\\ &0=0+0\\ &0=-4+(-2y-2x)\\ \end{cases}\\ &\text{The first two are identities, from the third we obtain the equation of the central axis}\\ &y=-x-2 \end{aligned}$
The above system is equivalent in terms of the effects of the operation to the system below

Statics - Planar force system

Engineering mechanics - Statics - Planar force system - Example 1

Example 1