To solve this problem, we can use properties of determinants. Specifically, we can use the linearity property of determinants, which states that if a row (or column) of a determinant is a linear combination of other rows (or columns), the determinant can be expressed as a sum of determinants. Here, the second row of the new matrix is a linear combination of the original second and third rows. We can express the determinant of the new matrix as a sum of two determinants: one with the second row as \(3(d, e, f)\) and the other with the second row as \((g, h, i)\). The first determinant is simply 3 times the original determinant, and the second determinant is zero because it has two identical rows.
We start with the determinant of the original matrix given by
\[
\left|\begin{array}{lll}a & b & c \\ d & e & f \\ g & h & i\end{array}\right| = 8.
\]
We need to find the determinant of the new matrix
\[
\left|\begin{array}{rrr}a & b & c \\ 3d + g & 3e + h & 3f + i \\ g & h & i\end{array}\right|.
\]
Using the linearity property of determinants, we can express this determinant as:
\[
\left|\begin{array}{rrr}a & b & c \\ 3d & 3e & 3f \\ g & h & i\end{array}\right| + \left|\begin{array}{rrr}a & b & c \\ g & h & i \\ g & h & i\end{array}\right|.
\]
The first determinant can be simplified as follows:
\[
\left|\begin{array}{rrr}a & b & c \\ 3d & 3e & 3f \\ g & h & i\end{array}\right| = 3 \left|\begin{array}{rrr}a & b & c \\ d & e & f \\ g & h & i\end{array}\right| = 3 \times 8 = 24.
\]
The second determinant is zero because it has two identical rows:
\[
\left|\begin{array}{rrr}a & b & c \\ g & h & i \\ g & h & i\end{array}\right| = 0.
\]
Thus, the determinant of the new matrix is:
\[
\left|\begin{array}{rrr}a & b & c \\ 3d + g & 3e + h & 3f + i \\ g & h & i\end{array}\right| = 24 + 0 = 24.
\]
The final result is
\[
\boxed{24}.
\]