5. In Pascal's Triangle, each number is the sum of the two numbers directly above it. The first few rows of this triangle are as follows:
\begin{tabular}{|c|c|c|c|c|c|c|}
\hline Row 0 & \multicolumn{6}{|c|}{1} \\
\hline Row 1 & & & 1 & & 1 & \\
\hline Row 2 & & 1 & & 2 & & 1 \\
\hline Row 3 & & & 3 & & 3 & 1 \\
\hline Row 4 & 1 & 4 & & 6 & & 4 \\
\hline
\end{tabular}

Then, in Pascal's Triangle, three consecutive numbers with a ratio of $3: 4: 5$ will appear in the $\qquad$ row.
5. Line 62.

In Pascal's Triangle, its $n$-th row consists of binomial coefficients $C_{n}^{k}$ $(k=0,1, \cdots, n)$. If the ratio of three consecutive terms in the $n$-th row of Pascal's Triangle is $3: 4: 5$, then there exists a positive integer $k$ such that
$$
\begin{array}{l}
\frac{3}{4}=\frac{C_{n}^{k-1}}{C_{n}^{k}}=\frac{k!(n-k)!}{(k-1)!(n-k+1)!}=\frac{k}{n-k+1}, \\
\frac{4}{5}=\frac{C_{n}^{k}}{C_{n}^{k+1}}=\frac{(k+1)!(n-k-1)!}{k!(n-k)!}=\frac{k+1}{n-k} .
\end{array}
$$

This leads to $\left\{\begin{array}{l}3 n-7 k=-3, \\ 4 n-9 k=5 .\end{array}\right.$
Solving the system of equations yields $k=27, n=62$.