Step 1 :Let the camera position be \(C(2, 2)\), the item position be \(I(3, -7)\), and the boundary vectors be \(c_{1} = -2i - 3j\) and \(c_{2} = 2i - j\).
Step 2 :Find the vector from the camera to the item: \(CI = I - C = (1, -9)\).
Step 3 :Find the vector projection of \(CI\) onto \(c_{1}\) and \(c_{2}\):
Step 4 :\(proj_{c_{1}}(CI) = \frac{CI \cdot c_{1}}{c_{1} \cdot c_{1}} c_{1} = \frac{(1, -9) \cdot (-2, -3)}{(-2, -3) \cdot (-2, -3)} (-2, -3) = (-3.84615385, -5.76923077)\)
Step 5 :\(proj_{c_{2}}(CI) = \frac{CI \cdot c_{2}}{c_{2} \cdot c_{2}} c_{2} = \frac{(1, -9) \cdot (2, -1)}{(2, -1) \cdot (2, -1)} (2, -1) = (4.4, -2.2)\)
Step 6 :Choose the direction vector that minimizes the distance traveled to get beyond the detection zone: \(\boxed{(4.4, -2.2)}\)