Final answer:
To find the required plane, calculate the cross product of the normals of the given planes to determine the direction vector of the line of intersection, then use the point and this direction vector to write the plane equation.
Step-by-step explanation:
To find the plane that passes through the point (-1, 1, 2) and contains the line of intersection of the planes x + y - z = 3 and 2x - y + 3z = 2, we first need to find the direction vector of the line of intersection. This can be done by taking the cross product of the normal vectors of the given planes.
The normal vector of the first plane is <1, 1, -1> and for the second plane is <2, -1, 3>. The cross product of these two vectors will give us the direction vector d of the line of intersection.
Next, we'll use the point (-1, 1, 2) and the direction vector d to write the equation of the plane. If d = , the equation of the plane can be written as:
A(x + 1) + B(y - 1) + C(z - 2) = 0, where A, B, and C are the components of d. The values for A, B, and C are obtained by plugging in the coordinates of d into the plane equation.
The final step is to simplify and write the equation of the plane in the general form Ax + By + Cz + D = 0.