To find the equation of the plane passing through a given point and perpendicular to a given line, we can use the following steps:

1. Find the direction vector of the line:
- The direction vector of a line passing through two points A(x1, y1, z1) and B(x2, y2, z2) is given by the difference between the coordinates of the two points: AB = (x2 - x1, y2 - y1, z2 - z1).

2. Find a normal vector to the plane:
- Since the plane is perpendicular to the line, the direction vector of the line is also a normal vector to the plane.

3. Use the point-normal form of the equation of a plane:
- The equation of a plane passing through a given point P(x0, y0, z0) and with a normal vector N(a, b, c) is given by the equation: N · (P - P0) = 0, where · denotes the dot product and P0 is the given point.

Now, let's apply these steps to the given problem:

1. Finding the direction vector of the line:
- The line passes through the points A(3, 4, -1) and B(2, -1, 5), so the direction vector AB is given by AB = (2 - 3, -1 - 4, 5 - (-1)) = (-1, -5, 6).

2. Finding a normal vector to the plane:
- Since the plane is perpendicular to the line, the direction vector of the line AB is also a normal vector to the plane. Therefore, the normal vector to the plane is N = (-1, -5, 6).

3. Using the point-normal form of the equation of a plane:
- We are given a point P(2, -3, 1) that lies on the plane. Let's substitute these values into the equation N · (P - P0) = 0:
(-1, -5, 6) · ((2, -3, 1) - (2, -3, 1)) = 0
(-1, -5, 6) · (0, 0, 0) = 0
0 = 0

The equation simplifies to 0 = 0, which is always true. This indicates that the given point P lies on the plane. Therefore, the equation of the plane passing through (2, -3, 1) and perpendicular to the line joining (3, 4, -1) and (2, -1, 5) is satisfied by all points in 3D space. The correct answer is option 'B' (x + 5y - 6z + 19 = 0).