To reduce the force system to pass through given point, we have to find resultant of force system and resultant moment about the given point.

Here we have to reduce the force system to force acting through origin and Moment acting about origin O.

Resultant of Forces :

From the given figure we can write Forces F1, F2, F3 in component form.

as shown below.

$\vec{F1}=200 \left(\frac{5}{\sqrt{34}}\hat{i} + \frac{-3}{\sqrt{34}}\hat{k}\right) $N

$\vec{F2} =100\hat{j}$ N

$\vec{F3} =400\left(\frac{5}{\sqrt{41}}\hat{i}+\frac{4}{\sqrt{41}}\hat{j}\right)$N

Hence, resultant force of the above forces will be

$\vec{F} =\vec{F1}+\vec{F2}+\vec{F3}\\ = 200 \left(\frac{5}{\sqrt{34}}\hat{i} + \frac{-3}{\sqrt{34}}\hat{k}\right)+ 100\hat{j} + 400\left(\frac{5}{\sqrt{41}}\hat{i}+\frac{4}{\sqrt{41}}\hat{j}\right)\\ =\left(\frac{1000}{\sqrt{34}}+\frac{2000}{\sqrt{41}}\right)\hat{i}+ \left( 100+\frac{1600}{\sqrt{41}}\right)\hat{j} + \frac{-600}{\sqrt{34}}\hat{k}\\ =(483.85\hat{i}+349.88\hat{j}-102.89\hat{k} ) N$

i.e Resultant force of magnitude

$|\vec{F}|=\sqrt{483.85^2+349.88^2+102.89^2}=605.89$N

acts in direction given by unit vector

$\frac{\vec{F}}{|\vec{F}|}= 0.7986\hat{i}+0.5775\hat{j}-0.1698\hat{k}$

Resultant of Moments :

Also, position vector of these force vectors are

$\vec{r1}=5\left(\frac{4}{5}\hat{j}+\frac{3}{5}\hat{k}\right)$

$\vec{r2}=5\hat{i}$

$\vec{r3}=3\hat{k}$

Hence Moment of these forces about origin are given as

$\vec{M1} = \vec{r1}\times \vec{F1} \\ =5\left(\frac{4}{5}\hat{j}+\frac{3}{5}\hat{k}\right) \times 200 \left(\frac{5}{\sqrt{34}}\hat{i} + \frac{-3}{\sqrt{34}}\hat{k}\right)\\ =1000 \left( \frac{-12}{5\sqrt{34}}\hat{i} + \frac{3}{\sqrt{34}}\hat{j} + \frac{-4}{\sqrt{34}}\hat{k} \right) Nm\\$ $\vec{M2} = \vec{r2}\times \vec{F2} \ =5\hat{i} \times 100\hat{j} \ = 500\hat{k} Nm$ $\vec{M3} = \vec{r3}\times \vec{F3} \ = 3\hat{k} \times 400\left(\frac{5}{\sqrt{41}}\hat{i}+\frac{4}{\sqrt{41}}\hat{j}\right) \ = 1200 \left( \frac{-4}{\sqrt{41}}\hat{i} + \frac{5}{\sqrt{41}} \hat{j} \right) Nm$ Hence Resultant moments will be $\vec{M} = \vec{M1}+\vec{M2}+\vec{M3} \ = 1000 \left( \frac{-12}{5\sqrt{34}}\hat{i} + \frac{3}{\sqrt{34}}\hat{j} + \frac{-4}{\sqrt{34}}\hat{k} \right) + 500\hat{k} + 1200 \left( \frac{-4}{\sqrt{41}}\hat{i} + \frac{5}{\sqrt{41}}\hat{j} \right) \ = \left( \frac{-2400}{\sqrt{34}}+ \frac{-4800}{\sqrt{41}}\right)\hat{i} + \left( \frac{3000}{\sqrt{34}}+ \frac{6000}{\sqrt{41}}\right)\hat{j} + \left( \frac{-4000}{\sqrt{34}}+ 500\right)\hat{k} \ =( -1161.23\hat{i}+ 1451.54\hat{j} -185.99\hat{k})Nm$ i.e Resultant Moment of magnitude $|\vec{M}|=\sqrt{1161.23^2+1451.54^2+185.99^2}=1868.16 Nm$ acts in direction given by unit vector $\frac{\vec{M}}{|\vec{M}|}= -0.6216\hat{i}+0.7769\hat{j}-0.0996\hat{k}$   **Hence reduced system of forces passing through origin is given as** **Force F with magnitude ​605.89 N acting in direction $(0.7986\hat{i}+0.5775\hat{j}-0.1698\hat{k})$** **And Moment M with magnitude 1868.16 Nm acting in direction $(-0.6216\hat{i}+0.7769\hat{j}-0.0996\hat{k})$**