Step 1: Write the given information.
$\vec{F} = 2\hat{i}+2\hat{j}+5\hat{k}$, $A=(1,2,5)$, $B=(-1,-2,-3)$. We are told $\overrightarrow{BA} \times \vec{F} = 4\hat{i}+6\hat{j}+2\lambda\hat{k}$.
Step 2: Find $\overrightarrow{BA}$.
$\overrightarrow{BA} = A - B = (1-(-1))\hat{i}+(2-(-2))\hat{j}+(5-(-3))\hat{k} = 2\hat{i}+4\hat{j}+8\hat{k}$.
Step 3: Compute $\overrightarrow{BA} \times \vec{F}$ using the determinant formula.
$\overrightarrow{BA} \times \vec{F} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 4 & 8 \\ 2 & 2 & 5 \end{vmatrix}$.
Step 4: Expand the determinant.
$\hat{i}(4 \cdot 5 - 8 \cdot 2) - \hat{j}(2 \cdot 5 - 8 \cdot 2) + \hat{k}(2 \cdot 2 - 4 \cdot 2)$
$= \hat{i}(20-16) - \hat{j}(10-16) + \hat{k}(4-8)$
$= 4\hat{i} - (-6)\hat{j} + (-4)\hat{k}$
$= 4\hat{i} + 6\hat{j} - 4\hat{k}$.
Step 5: Compare with the given result.
Given: $4\hat{i}+6\hat{j}+2\lambda\hat{k}$. Comparing the $\hat{k}$ component: $2\lambda = -4$, so $\lambda = -2$.
Step 6: Match with options.
$\lambda = -2$ is option (4).
\[ \boxed{\lambda = -2} \]