How many nodal planes are there in atomic orbitals corresponding to the principal quantum number $n=3$?

The correct option is B 11
For $n=3$, values of $l$ corresponds to $\left(0to2\right)$. The corresponding orbitals are as follows:
$3s-\text{orbital}3s$
$3p-\text{orbital}3p_x,3p_y,3p_z$
$3d-\text{orbital}3d_{xy},3d_{yz},3d_{zx},3d_{x^2-y^2},3d_{z^2}$
We know that,
'$s$' orbital has no nodal plane.
Each of $p_x,p_y,p_z$ has one nodal plane, which means a total of three nodal planes.
Each of $d_{xy},d_{yz},d_{zx},d_{x^2-y^2}$ has two nodal planes each and $d_{z^2}$ has no nodal plane which means a total of eight nodal planes.

Hence, for $n=3$, there are a total of $11$ nodal planes.