This example uses count to count the number of times 0 appears among v1–v3 (putting it in z1) and likewise for v4–v6 (putting the count in z2). Then, if z1 is 3, then 9999 is placed in v1–v3 using a do repeat loop. The same is done for v4–v6.
data list free / v1 to v6. begin data. 0 0 0 1 0 1 1 1 1 1 1 0 0 1 1 0 0 0 0 0 0 0 0 0 end data. count z1=v1 to v3 (0). count z2=v4 to v6 (0). do repeat x=v1 to v3. if z1=3 x=9999. end repeat. do repeat x=v4 to v6. if z2=3 x=9999. end repeat. execute.