Step-by-step Solution:

Given that \( \vec a + 2 \vec b \) is collinear with \( \vec c \), we have: \[ \vec a + 2 \vec b = \lambda \vec c \quad \text{(1)} \] Also, \( \vec b + 3 \vec c \) is collinear with \( \vec a \), so: \[ \vec b + 3 \vec c = \mu \vec a \quad \text{(2)} \] Equating the value of \( 2 \vec b \) from the above two equations, we get: \[ \lambda \vec c - \vec a = 2 \mu \vec a - 6 \vec c \] By comparison, we find that \( \lambda = -6 \). Substituting this value into the first equation, we get: \[ \vec a + 2 \vec b = -6 \vec c \] Or: \[ \left| \vec a + 2 \vec b + 6 \vec c \right| = 0 \]