To find the equation of line \( L \) that is perpendicular to the given plane and passes through the point \((-1, 2, -3)\), we need to follow these steps:

1. Understanding Perpendicularity: A line perpendicular to a plane will have a direction vector that is parallel to the normal vector of the plane. The equation of the plane is \(2x + 3y + z + 5 = 0\). The normal vector to this plane is \(\mathbf{n} = \langle 2, 3, 1 \rangle\).
2. Point on the Line: The line \( L \) passes through the point \((-1, 2, -3)\). Therefore, the line can be represented parametrically as:
   • \(x = -1 + 2t\)
   • \(y = 2 + 3t\)
   • \(z = -3 + t\)
3. Eliminating the Parameter: To form symmetric equations, eliminate \( t \):
   • From \( x = -1 + 2t \), we rearrange to get \( t = \frac{x + 1}{2} \).
   • From \( y = 2 + 3t \), we rearrange to get \( t = \frac{y - 2}{3} \).
   • From \( z = -3 + t \), we rearrange to get \( t = z + 3 \).
4. Matching with Options: The symmetric form matches with:

\(- (x + 1) = y - 2 = z + 3\)

Thus, the correct answer is that the equation of line \( L \) is

\(- (x + 1) = y - 2 = z + 3\)