So A hits the target 1/3 of the time, B hits it 2/5 of the time, and C hits it 1/2 of the time. We can, perhaps by using a "Venn diagram", divide this into 7 "mutually exclusive" cases:
A, B, and C all hit the target
A and B hit the target and C misses.
A and C hit the target and B misses.
B and C hit the target and A misses.
A hits the target and B and C miss.
B hits the target and A and C miss.
C hits the target and A and B miss.
The question asks for the probability that IF only one person hits that person is A.
Of course, only A hitting the target is precisely
"A hits the target and B and C miss".
The probability A hits the target is given as 1/3 while the probability B misses is 1- 2/5= 3/5 and the probability C misses is 1- 1/2= 1/2. What is the probability A hits while B and C miss?
But we want the probability this happens given that exactly one of the three hits. We need to divide by the probability exactly one hits which is the sum of the probabilities
"A hits the target and B and C miss" (this is, of course, the same as calculated above)
"B hits the target and A and C miss"
"C hits the target and A and B miss".
Calculate the probability of each of those and add them. Then divide the first probability, above, by that sum.