Let T be the orthogonal projection of S onto the plane containing PQR. Let U be the point on the line containing QR such that TU is perpendicular to QR.
Then STU is a right triangle. It lies on a plane perpendicular to line QR. Then ST is the height of the tetrahedron relative to face PQR, angle STU is a right angle, and angle SUT is the dihedral angle between faces PQR and QRS.
SU equals the altitude of QRS perpendicular to QR, which equals 2*area(QRS)/QR. ST equals SU*sin(angle SUT). Then the volume of the tetrahedron PQRS is given by Volume = (1/3)*area(PRQ)*(2*area(QRS)/QR)*sin(angle SUT)
This simplifies to Volume = (2/3) * area(PRQ) * area(QRS) * sin(dihedral angle) / QR
Then plug in the given values: Volume = (2/3) * 80 * 120 * sin(30 deg) / 10 = 320
Solution
