Final answer:
The equation of a plane passing through the point (3, -1, 3) and perpendicular to the vector (6, 5, 4) is 6x + 5y + 4z - 35 = 0.
Step-by-step explanation:
To find an equation of a plane that passes through a given point (3, -1, 3) and is perpendicular to a given vector (6, 5, 4), we use the fact that the normal vector of the plane (which is perpendicular to the plane) will have the same components as the given vector.
The general equation of a plane in three-dimensional space is Ax + By + Cz + D = 0, where A, B, and C are the components of the normal vector to the plane, and D is a constant. Plugging in the components of the given vector for A, B, and C, we get the equation 6x + 5y + 4z + D = 0.
We determine D by plugging in the coordinates of the given point into the equation, resulting in 6(3) + 5(-1) + 4(3) + D = 0. Solving for D gives us D = -35.
Therefore, the equation of the plane is 6x + 5y + 4z - 35 = 0.