The correct option is A →a and →c are like but →b and →c are unlike vectors
Given →a+5→b=→c and →a−7→b=2→c
Eliminating →b gives
→c−→a5=→a−2→c7
⇒7→c−7→a=5→a−10→c
⇒12→a=17→c -----(1)
Eliminating →a gives
→c−5→b=2→c+7→b
→c=−12→b -----(2)
∴ From (1) and (2)
→a and →c are like but →b and →c are unlike vectors.
Hence, option A is correct.