Each person might be allotted more than one share of the secret.
Let $G$, $C$ and $D$ denote the number of shares allotted to a General, a Colonel, and a Desk Clerk respectively, and let $T$ denote the Threshold of the secret sharing scheme. Then, we have that
T &\leq 5G,\\
T &\leq 4C + 3D,\\
T &\leq 3G + 3D.
Can you find suitable values for $G,C,D$,and $T$? The total number of shares is
$N = 6G + 5C + 4D$ and we have a $(T,N)$ secret sharing scheme.
$D=2, C=3, G=4, T=18$, and $N = 47$ seems to work. There are, of course, other combinations that would result in $18$ or more shares, e.g., $5C+2D$ or $4G+D$ but nothing in the problem statement says that this is not to be allowed. Note that the $5$ colonels (or the $4$ Desk Clerks for that matter) cannot stage a coup by themselves; the colonels have to have at least two Desk Clerks (or a General) as co-conspirators.
Addendum: in the spirit of @IlmariKaronen's answer,
Divide the secret $S$ into 6 shares in a $(5,6)$ secret-sharing scheme and give each of the six generals a share. Any 5 of them can reconstruct $S$.
Create a random binary vector $X$ as long as the secret $S$, and then make five shares of $S\oplus X$ in a $(4,5)$ secret-sharing scheme, giving one share to each colonel. Any four of them can reconstruct $S\oplus X$. Create $4$ shares of $X$ in a $(3,4)$ secret-sharing scheme and hand one share to each desk clerk. Any $3$ desk clerks can reconstruct $X$, and together with $S\oplus X$ from the $4$ colonels, $S$ can be reconstructed.
Create $6$ shares of $S\oplus X$ in a $(3,6)$ secret-sharing scheme and hand a share to each general. Any three generals can reconstruct $S\oplus X$, and together with $X$ from three desk clerks, can recreate $S$.
Note that this has have fewer secret-sharing schemes to implement than Ilmari's method, and the desk clerks and colonels have only one share to have and hold. Only the generals have two different shares and must remember to use the correct one in the two different cases when they are acting by themselves and when in conjunction with three desk clerks. Also, the desk clerks, by themselves, can only reconstruct $X$.