Condition the multi-pack card sample on its suit counts

The grand total 40 completes the source row/column totals. Physical copies are distinct in the random sampling, even when their suit/rank labels match.

The all-King, all-Queen and all-Jack events are disjoint: $P=(15/40)(14/39)(13/38)+(14/40)(13/39)(12/38)+(11/40)(10/39)(9/38)=123/1235$.

$P=[\binom{15}3+\binom{14}3+\binom{11}3]/\binom{40}3=(455+364+165)/9880=123/1235$.

Only King or Queen paths can qualify: there is only one Jack of Spades, so two Jack Spades are impossible. In one suit order the two King Spades/King Diamond path is $(3/40)(2/39)(6/38)$, and the Queen path is $(4/40)(3/39)(4/38)$. Each has $3!/2!$ suit orders. The conditioning event has probability $(8/40)(7/39)(12/38)(3!/2!)$. Thus

$\begin{aligned}&P\\ & =\frac{(3/40)(2/39)(6/38)(3!/2!)+(4/40)(3/39)(4/38)(3!/2!)}{(8/40)(7/39)(12/38)(3!/2!)}\\ & =1/8.\end{aligned}$

The conditioned sample space has $\binom82\binom{12}1=336$ equally likely physical-card selections. The favourable ones are $\binom32\binom61=18$ King selections and $\binom42\binom41=24$ Queen selections. Hence $(18+24)/336=1/8$.