According to Handshaking-Lemma Sum of the degree of all the vertices is even.
that is $ \sum_{ i =1 }^n \text{degree} (v_i)=2e$
Note: Degree of vertices = the number of edges incident with it.
Given: 1) Number of vertices with $d(1)=2n$
2) Number of vertices with $d(2)=3n$
3) Number of vertices with $d(3)=n$
The number of edges is not given in question but we know that a tree $T$ having $n$ vertices contain $(n-1)$ edges.
Here total number of vertices $(V) =2n+3n+n=6n \qquad\dots\dots(i)$
Total number of edges $(E) = (6n-1) \qquad \dots\dots (ii)$
$\because$ Sum of the degree of all vertices $=2*$ number of edges
$\implies 2n*1+3n*2+n*3= 2*(6n-1)$
$\implies 2n+6n+3n=12n-2$
$\implies 11n=12n-2$
$\implies n=2$
Put $n=2$ in eq $(i),(ii)$,we get
Total number of vertices $(V)= 6*2=12$
Total number of edges $(E)= 6*2-1=11$
Option $(A)$ is correct.
Ref:1) Gate 2004
2) Gate 2017