vector AB = (-1,-1,-6)
vector AC = (-1,5,-5)
a normal to these two vectors is (35, 1, -6) , I assume you know how to take a cross-product
the plane has equation
35x+ y - 6z = c
plug in a point, e.g. A(6,2,3)
210 + 2 - 18 = c
plane equation: 35x + y - 6z = 194
check by subbing in points B and C, (they work)
b) Since they are parallel, they must differ only in the constant
so new equation: 2x + 6y + 4z = c
plug in your point to find c
1. Find vectors AB and AC.
2. Calculate the cross product of ABxAC, which gives the normal vector to the required plane.
3. Form the equation of the plane. The components of the normal vector correspond to the coefficients of x, y and z of the equation of the plane.
4. Find the constant term of the plane by passing the plane through one of the three points, i.e. substitute (xa,ya,za) into the equation and find the constant term.
5. Optionally, check that the other points pass through the equation of the plane.
Parallel to 2x + 6y + 4z = 1 means that the normal is <2,6,4>.
Take the left-hand side of the equation and evaluate with the point P(3,2,1), i.e. pass a plane parallel to 2x + 6y + 4z = 1 through the given point:
So the required equation of the plane is
2x + 6y + 4z = 16