Final answer:
The parametric equations of the line passing through the point (-2, 1, 3) that is perpendicular to the plane with the equation 2x - 3y = 7 are x = -2 + 3t, y = 1 + 2t, and z = 3 - t.
Step-by-step explanation:
To find the parametric equations of the line passing through a given point and perpendicular to a plane, we can use the normal vector of the plane. The given plane has the equation 2x - 3y = 7, and its normal vector is ⟨2, -3, 0⟩.
This vector becomes the direction vector of the line we seek. To parametrize the line, we use the point-direction form of a line equation: x = x₀ + at, y = y₀ + bt, z = z₀ + ct, where (x₀, y₀, z₀) is the given point, and ⟨a, b, c⟩ is the direction vector. Substituting the values, we get the parametric equations x = -2 + 3t, y = 1 + 2t, and z = 3 - t.
These equations represent the line passing through the given point and perpendicular to the specified plane.