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It is given that,

$$\begin{gathered} \stackrel{鈫�}{v}=2\hat{i}-7\hat{j} \\ {\stackrel{鈫�}{\mathrm{F}}}_1=2\hat{\mathrm{i}}+3\hat{\mathrm{j}}-2\hat{\mathrm{k}} \\ {\stackrel{鈫�}{\mathrm{F}}}_2=-5\hat{\mathrm{i}}+8\hat{\mathrm{j}}-2\hat{\mathrm{k}} \end{gathered}$$

The problem is based on Newton鈥檚 second law of motion which states that the rate of change of momentum of a body is equal in both magnitude and direction of the force acting on it.

Formula:

According to Newton鈥檚 second law,

$\underset{}{鈭�}{F}_{net}=ma$

where,$F_{net}$ is the net force, mis mass and a is an acceleration.

If velocity is constant, acceleration would be zero.According to Newton鈥檚 second law, acceleration would be zero when net force is zero.

Thus,

$$\begin{gathered} \underset{}{鈭�}{\stackrel{鈫�}{F}}_{net}=0 \\ {\stackrel{鈫�}{F}}_1+{\stackrel{鈫�}{F}}_2+{\stackrel{鈫�}{F}}_3=0 \end{gathered}$$

Consider,$F_3={\mathrm{F}}_{\mathrm{x}}\hat{\mathrm{i}}+{\mathrm{F}}_{\mathrm{y}}\hat{\mathrm{j}}+{\mathrm{F}}_2\hat{\mathrm{k}}$

Hence,

role="math" localid="1660907302155" $0=\left(2-5+F_x\right)\hat{i}+\left(3+8+F_z\right)\hat{k}$

Now, by comparing the coefficients of$\hat{\mathrm{i}},\hat{\mathrm{j}}\mathrm{and}\hat{\mathrm{k}}$on both sides,

$$\begin{gathered} F_x=3 \\ {F}_y=-11 \\ {F}_z=4 \end{gathered}$$

Hence, the unknown force $F_3$acting on the particle is$F_x=3\hat{\mathrm{i}}-11\hat{\mathrm{j}}+4\hat{\mathrm{k}}$