Find the equation of the plane through the point p=(3,2,3),and paralled to the plane y-x+2z = -1.

Step 1: Understand the problem We are asked to find the equation of a plane that passes through a given point and is parallel to a given plane.

Step 2: Recall the equation of a plane The general equation of a plane in 3D space is given by Ax + By + Cz = D, where A, B, C, and D are constants, and x, y, and z are the coordinates of any point on the plane.

Step 3: Understand parallel planes Two planes are parallel if and only if their normal vectors are parallel. The normal vector of a plane is the vector that is perpendicular to the plane. In the equation of a plane Ax + By + Cz = D, the normal vector is (A, B, C).

Step 4: Find the normal vector of the given plane The given plane is y - x + 2z = -1. Comparing this with the general equation of a plane, we can see that the normal vector of the given plane is (-1, 1, 2).

Step 5: Use the normal vector for the new plane Since the new plane is parallel to the given plane, their normal vectors are the same. Therefore, the normal vector of the new plane is also (-1, 1, 2).

Step 6: Use the given point to find the equation of the new plane The new plane passes through the point p = (3, 2, 3). We can substitute these coordinates into the general equation of a plane to find the value of D.

-13 + 12 + 2*3 = D -3 + 2 + 6 = D D = 5

Step 7: Write the equation of the new plane Substituting A = -1, B = 1, C = 2, and D = 5 into the general equation of a plane, we get the equation of the new plane:

-1x + 1y + 2z = 5

Or, simplifying, we get:

x - y + 2z = -5

This is the equation of the plane that passes through the point p = (3, 2, 3) and is parallel to the plane y - x + 2z = -1.