To find the equation of the plane passing through the point (2, 3, -1) and perpendicular to the line with direction ratios 1, -2, and 3, we can use the formula for the equation of a plane:

The equation of a plane is given by:

A(x – x₀) + B(y – y₀) + C(z – z₀) = 0

Where (x₀, y₀, z₀) is a point on the plane, and A, B, and C are the direction ratios of the normal vector to the plane.

Since the plane is perpendicular to the line with direction ratios 1, -2, and 3, the direction ratios of the line give the normal vector to the plane. Therefore, A = 1, B = -2, and C = 3.

Now substitute the point (2, 3, -1) and the values of A, B, and C into the equation:

1(x – 2) – 2(y – 3) + 3(z + 1) = 0

Simplify the equation:

(x – 2) – 2(y – 3) + 3(z + 1) = 0

x – 2 – 2y + 6 + 3z + 3 = 0

x – 2y + 3z + 7 = 0

Thus, the equation of the plane is:

x – 2y + 3z + 7 = 0