The vector normal to the plane is (1,1,2). You know it from the coefficients in front of x,y and z.
You know that this vector will be part of the perpendicular plane since it is the normal vector. Then you want the vector linking points (2,-1,4) , (3,2,-1) to be in the perpendicular plane as well.
Find this vector and compute the cross product with the normal vector (1,1,2). This will give you the normal vector of the perpendicular plane (say (a,b,c)).
The equation of the plane is where are the point of either of the points (2,-1,4) , (3,2,-1).