Find the equation of the plane that is parallel to the line of intersection of the planes with equations $2x-3y+z=5$ and $x+y-z=2$, and passes through the points A(1,2,3), B(2,-1,1), C(-1,2,2) and D(0,1,-1).

Step 1 ：First find the direction ratios of the line of intersection of the given planes. This can be found by taking the cross product of the normal vectors of the two planes. The normal vector of the first plane is $2\hat{i}-3\hat{j}+\hat{k}$ and of the second plane is $\hat{i}+\hat{j}-\hat{k}$. So, the cross product is $-3\hat{i}-3\hat{j}+5\hat{k}$.

Step 2 ：The equation of the plane passing through three non-collinear points A, B and C can be found by first finding the vectors AB and AC. $\overrightarrow{AB}=\overrightarrow{j}-3\overrightarrow{k}$ and $\overrightarrow{AC}=-2\overrightarrow{i}+\overrightarrow{j}-\overrightarrow{k}$. The cross product of these two vectors gives the normal to the plane. $\overrightarrow{AB}\times \overrightarrow{AC}=-2\overrightarrow{i}-3\overrightarrow{j}+5\overrightarrow{k}$.

Step 3 ：These normals are parallel, so the planes are parallel. The equation of the plane can now be found using the equation of a plane in normal form, $n\cdot (x-x_0,y-y_0,z-z_0)=0$, where $(x_0,y_0,z_0)$ is a point on the plane and n is the normal vector. Substituting the coordinates of point A and the normal vector, we get $-2(x-1)-3(y-2)+5(z-3)=0$. Simplifying this gives $-2x-3y+5z=-1$.