Final answer:
The equation of the plane that passes through the point Po(-3,1,7) and is perpendicular to the given line is 6x - 5y + 4z = 5.
Step-by-step explanation:
The question involves finding the equation of a plane that is perpendicular to a given line and passes through a specific point. This type of problem is solved by using the vector that is perpendicular to the plane as the normal vector in the plane equation.
To determine the equation of the plane, one can apply the general plane equation Ax + By + Cz = D, where A, B, and C are the components of the normal vector to the plane. Given the line equations x= -1 + 6t, y = 2 - 5t, z = 3 + 4t, the coefficients of t (6, -5, 4) are the components of the direction vector of the line, which is also the normal vector of the plane.
Using the point Po(-3,1,7), we substitute the normal vector and the point into the plane equation to get 6(-3) - 5(1) + 4(7) = D. Solving for D gives us D = 5. Hence, the equation of the plane is 6x - 5y + 4z = 5.