Let "h" be height of the 3-ary tree.
Since we know that Leaves are present at last level (in case of maximum number of leaves like in Complete 3-ary tree).
So, number of Leaves (say L) will be "atmost" = 3^{h}.
ex- we can have a maximum of L=1 at h=0 (when root itself is leaf)
L=3 at h=1
L=9 at h=2 and so on.^{ }
Similarly, Internal nodes are present in maximum at (h-1) height.
So,Number of Internal Nodes (say I) will be "atmost "
=3^{0}+3^{1}+3^{2}+.........3^{h-1 }= 3^{(h -1)}/2
Let n be Total number of nodes, then n= L+I
n= 3^{h} + 3^{h -1}/2
n= 3^{h+1} -1/2
2n+1=3.3^{h}
2n+1=3.L ( since L= 3^{h)}
hence L= (2n+1)/3