Hello Noelia,
First, let's draw the Venn diagram:
View attachment 1144
The set $B$ represents those students who got a certificate in biology, $C$ represents those students who got a certificate in chemistry, and $P$ represents those students who got a certificate in physics.
We are told the cardinality of set $B$ is 14, so we may write:
$$a+x+y+3=14\implies a=11-(x+y)$$
We are told the cardinality of set $C$ is 13, so we may write:
$$b+y+z+3=13\implies b=10-(y+z))$$
We are told the cardinality of set $P$ is 21, we we may write:
$$c+x+z+3=21\implies c=18-(x+z))$$
Hence, the number of students that received only 1 certificate is:
$$a+b+c=39-2(x+y+z)$$
The number of students that received 2 certificates is:
$$x+y+z$$
And we are told the the number of students that received 3 certificates is:
$$3$$
The sum of these is 34, since we are told every student received at least one certificate:
$$39-2(x+y+z)+x+y+z+3=34$$
$$x+y+z=8$$
And so:
$$a+b+c=39-2(8)=23$$
And so we may conclude that 3 students received 3 certificates, 8 students received 2 certificates, and 23 received 1 certificate.
