Pascal row with neighbouring entries in ratio 3 to 4 to 5

Some row $n$ of Pascal's triangle contains three adjacent entries whose ratio is $3:4:5$. What is the smallest such $n$?

$62$. Write the entries as $\binom{n}{k-1},\binom nk,\binom n{k+1}$. Their consecutive ratios are $\tfrac{k}{n-k+1}=\tfrac34$ and $\tfrac{k+1}{n-k}=\tfrac45$, i.e. $7k=3n+3$ and $9k=4n-5$, whose unique solution is $n=62$, $k=27$.