To solve this problem, we will use the formula for the magnetic force acting on a current-carrying wire in a magnetic field.$\\$The magnetic force $\vec{F}$ on a wire of length $L$ carrying a current $I$ in a magnetic field $\vec{B}$ is given by:$\\ \vec{F} = I (\vec{L} \times \vec{B}) \\$Here, the wire is along the positive x-axis, so the length vector $\vec{L}$ can be expressed as:$\\ \vec{L} = L \hat{i} \\$The magnetic field is given as:$\\ \vec{B} = 2\hat{i} + 3\hat{j} - 4\hat{k} \\$Now, calculate the cross product $\vec{L} \times \vec{B}: \\ \vec{L} \times \vec{B} = (L \hat{i}) \times (2\hat{i} + 3\hat{j} - 4\hat{k}) \\$Using the properties of cross products:$\\ \hat{i} \times \hat{i} = 0, \quad \hat{i} \times \hat{j} = \hat{k}, \quad \hat{i} \times \hat{k} = -\hat{j} \\$We have:$\\ \vec{L} \times \vec{B} = L (\hat{i} \times 2\hat{i} + \hat{i} \times 3\hat{j} + \hat{i} \times (-4\hat{k})) \\ = L (0 + 3\hat{k} + 4\hat{j}) \\ = L (4\hat{j} + 3\hat{k}) \\$Thus, the magnetic force $\vec{F}$ is:$\\ \vec{F} = I L (4\hat{j} + 3\hat{k}) \\$The magnitude of the magnetic force is given by:$\\ |\vec{F}| = I L \sqrt{4^2 + 3^2} \\ = I L \sqrt{16 + 9} \\ = I L \sqrt{25} \\ = 5 I L \\$Therefore, the magnitude of the magnetic force acting on the wire is $5 I L$. This corresponds to Option 3.