The plane equation is AX+BY+CZ+D=0 and its normal vector is n(A, b, c). (I'll take A as A( 1, 2,-1) for the time being. . )

Two vectors AB(0, 2, -3) and AC( 1,-1, 4) can be determined from three points A, B and C. According to geometric knowledge, the normal vectors are perpendicular to AB and AC vectors respectively. There are two ways to solve this problem.

Method 1: n is multiplied by vectors AB and AC respectively, and both are equal to 0. Form an equation group and solve the relationship between A, B and C, that is, use one to represent the other two (for example, use C to represent A and B). Finally, substitute point A to get the relationship between D and C, and then simplify the unknown to get the answer. Of course, in the end, you can also bring in two points and solve the equation directly.

Method 2: I am also a freshman. I wonder if you offer the course of linear algebra. If you do this, you will understand the second method well. Because n is perpendicular to AB and AC vectors, n=ab*ac=|i j k |

|0 -2 3 |=(-

| 1 - 1 4|

5,3,2). Substitute into point A and the equation can be solved.