Step 1 :Since the finish line coincides with the starting line, we have $\vec{A} + \vec{B} + \vec{C} + \vec{D} = \vec{0}$
Step 2 :Thus, $\vec{D} = - (\vec{A} + \vec{B} + \vec{C})$
Step 3 :Let $\alpha$ be the angle between $\vec{A}$ and $\vec{B}$, and $\beta$ be the angle between $\vec{B}$ and $\vec{C}$
Step 4 :Using the Law of Cosines, we have $D^2 = A^2 + B^2 - 2AB \cos \alpha + B^2 + C^2 - 2BC \cos \beta + A^2 + C^2 - 2AC \cos (\alpha + \beta)$
Step 5 :Substitute the given values, $D^2 = 3.50^2 + 4.60^2 - 2(3.50)(4.60) \cos 120^\circ + 4.60^2 + 5.10^2 - 2(4.60)(5.10) \cos 90^\circ + 3.50^2 + 5.10^2 - 2(3.50)(5.10) \cos 150^\circ$
Step 6 :Calculate the value of $D^2$, $D^2 = 3.50^2 + 4.60^2 - 2(3.50)(4.60) \cos 120^\circ + 4.60^2 + 5.10^2 + 3.50^2 + 5.10^2 - 2(3.50)(5.10) \cos 150^\circ$
Step 7 :Calculate the value of $D$, $D = \sqrt{D^2} = \boxed{4.20}$
Step 8 :Using the Law of Cosines to find $\theta$, we have $D^2 = A^2 + C^2 - 2AC \cos \theta$
Step 9 :Substitute the given values, $4.20^2 = 3.50^2 + 5.10^2 - 2(3.50)(5.10) \cos \theta$
Step 10 :Solve for $\cos \theta$, $\cos \theta = \frac{3.50^2 + 5.10^2 - 4.20^2}{2(3.50)(5.10)}$
Step 11 :Calculate the value of $\cos \theta$, $\cos \theta = \frac{3.50^2 + 5.10^2 - 4.20^2}{2(3.50)(5.10)}$
Step 12 :Find the value of $\theta$, $\theta = \arccos \left( \frac{3.50^2 + 5.10^2 - 4.20^2}{2(3.50)(5.10)} \right) = \boxed{75.0^\circ}$