Let's solve the problem step by step:

1. Identify the given information:
- PQRS is a parallelogram
- Side RS is produced to point T
- QT intersects side PS at point A
- AT = AQ = 6 cm
- AS = 3 cm
- TS = 4 cm

2. Determine the properties of a parallelogram:
- Opposite sides of a parallelogram are equal in length
- Opposite angles of a parallelogram are equal in measure
- Adjacent angles of a parallelogram are supplementary (their sum is 180 degrees)

3. Use the given information to find the value of x:
- Since AS = 3 cm and AT = 6 cm, we can conclude that ST = AT - AS = 6 cm - 3 cm = 3 cm
- In triangle ATS, we have AS = 3 cm, AT = 6 cm, and TS = 3 cm
- By applying the Pythagorean theorem, we can find the length of QS:
QS^2 = TS^2 + QT^2
QS^2 = 3^2 + x^2
QS^2 = 9 + x^2

4. Use the information about the parallelogram to find the value of y:
- Since AS = 3 cm and QS is a side of the parallelogram, we can conclude that PS = AS = 3 cm
- In triangle PSA, we have PS = 3 cm, AS = 3 cm, and angle PSA = angle AQS (opposite angles of a parallelogram)
- By applying the Law of Cosines, we can find the length of QS:
QS^2 = PS^2 + AS^2 - 2(PS)(AS)cos(angle PSA)
QS^2 = 3^2 + 3^2 - 2(3)(3)cos(angle AQS)
QS^2 = 18 - 18cos(angle AQS)

5. Equate the expressions for QS^2 from steps 3 and 4:
9 + x^2 = 18 - 18cos(angle AQS)

6. Solve for x:
x^2 + 18cos(angle AQS) = 9
x^2 = 9 - 18cos(angle AQS)
x^2 = 9 - 18(4/6) [Substituting the value of TS = 3 cm and AS = 3 cm into the equation]
x^2 = 9 - 12
x^2 = 9 - 12
x^2 = -3

Since the square of a real number cannot be negative, we cannot find a real value for x. Therefore, there is no solution for x in this problem.

7. Determine the value of y:
- Since AS = 3 cm and PS = 3 cm, we can conclude that QS = PS = 3 cm
- In triangle QTS, we have QT = 3 cm, TS = 3 cm, and QS = 3 cm
- By applying the Pythagorean theorem, we can find the length of QS:
QS^2 = QT^2 + TS