(1 point) The general (Cartesian) form for the plane passing through the point [math] and perpendicular to the line [math] with vector form [math] is .

It seems like you've missed to provide the specific point and line for the problem. However, I can explain the general process to find the Cartesian form of a plane passing through a given point and perpendicular to a given line.

1. Identify the given point and the direction vector of the line. The given point will be (x0, y0, z0) and the direction vector of the line will be (a, b, c).

2. Since the plane is perpendicular to the line, the direction vector of the line is also the normal vector of the plane.

3. The general form of the equation of a plane is Ax + By + Cz = D, where (A, B, C) is the normal vector of the plane.

4. Substitute the normal vector (a, b, c) into the equation to get ax + by + cz = D.

5. To find D, substitute the given point (x0, y0, z0) into the equation: ax0 + by0 + c*z0 = D.

6. The final equation of the plane is ax + by + cz = D, where D is the value obtained in step 5.

Please provide the specific point and line so I can give a more specific answer.