1. **State the problem:** We need to find the direction of the resultant vector $\vec{R}$ of three vectors $\vec{A}$, $\vec{B}$, and $\vec{C}$ with respect to the x-axis, denoted as $\theta_x$. 2. **Given vectors:** $$\vec{A} = 3\hat{i} + 7\hat{j} + 8\hat{k}$$ $$\vec{B} = 4\hat{i} - 5\hat{j} + 3\hat{k}$$ $$\vec{C} = 2\hat{i} + 3\hat{j} - 4\hat{k}$$ 3. **Formula for resultant vector:** $$\vec{R} = \vec{A} + \vec{B} + \vec{C}$$ 4. **Calculate the components of $\vec{R}$:** $$R_x = 3 + 4 + 2 = 9$$ $$R_y = 7 - 5 + 3 = 5$$ $$R_z = 8 + 3 - 4 = 7$$ 5. **Direction with x-axis:** The angle $\theta_x$ between $\vec{R}$ and the x-axis is given by $$\cos \theta_x = \frac{R_x}{|\vec{R}|}$$ where $$|\vec{R}| = \sqrt{R_x^2 + R_y^2 + R_z^2}$$ 6. **Calculate magnitude of $\vec{R}$:** $$|\vec{R}| = \sqrt{9^2 + 5^2 + 7^2} = \sqrt{81 + 25 + 49} = \sqrt{155}$$ 7. **Calculate $\cos \theta_x$:** $$\cos \theta_x = \frac{9}{\sqrt{155}}$$ 8. **Calculate $\theta_x$:** $$\theta_x = \cos^{-1} \left( \frac{9}{\sqrt{155}} \right)$$ 9. **Final answer:** $$\theta_x \approx \cos^{-1}(0.722) \approx 43.96^\circ$$ Thus, the direction of the resultant vector with the x-axis is approximately $43.96^\circ$.