There are infinitely many planes containing any two given points. To see this, visualise the line joining the points as the spine of a book, and the infinitely many planes as pages of the book.

The equation of the plane through the point (p,q,r) perpendicular to ai + bj + ck is a(x - p) + b(y - q) + c(z - r) = 0. Hence in this case, the Cartesian equation is 1(x - 3) - 1(y - 1) + 2(z - 0) = 0; that is, x - y + 2z = 2.

In order to know the equation of a plane, we must know a point on the plane and a vector perpendicular to the plane. We know three points on the plane, so we’re OK there. We can find a vector perpendicular to the plane by using the vector cross product. Notice that the vectors and are both parallel to the plane, so their cross product is normal to the plane. Since = -2j+6k and = -3i-2j+k, we calculate × = 10i-18j-6k. This is a normal vector to the plane. The cartesian equation is therefore 10(x - 1) - 18(y - 3) - 6(z + 1) = 0, or 10x - 18y - 6z = -38.

The two planes are obviously not parallel, since the first plane has normal vector n = 3i + j + k and the second plane has normal m = i - 2j + k, and n /|| m. Denoting the planes by ABCD and EBCF (with F obscured), then BC is the line of intersection.

Therefore

and from this we get radians.

The answer is 1.32 radians. The concept of “the angle between two planes” is somewhat ambiguous. For example, the line EB lies in one plane, and makes different angles with the lines AB, CB and DB in the other plane. To make it unambiguous, we define the angle between the two planes to be the angle between the normals n and m to the two planes.