498. Three automatic lines produce the same product, but have different productivity. The combined productivity of all three lines working simultaneously is 1.5 times the productivity of the first and second lines working simultaneously. A shift assignment for the first line can be completed by the second and third lines working simultaneously 4 hours and 48 minutes faster than the first line. The same assignment is completed by the second line 2 hours faster than the first line. How many hours does the first line take to complete its shift assignment?
$\triangle$ Let's accept the shift task for the first line as a unit.

Let the first line complete its shift task in $x$ hours, and the third line the same task in $y$ hours. Then the second line completes the task of the first in $(x-2)$ hours.

Therefore, the productivity of the first, second, and third lines are respectively $\frac{1}{x}, \frac{1}{x-2}$, and $\frac{1}{y}$ (in fractions of the first line's task per hour).

Since the combined productivity of all three lines is 1.5 times greater than the combined productivity of the first and second lines, then

$$
\frac{1}{x}+\frac{1}{x-2}+\frac{1}{y}=1.5\left(\frac{1}{x}+\frac{1}{x-2}\right)
$$

Since the second and third lines, working simultaneously, can complete the shift task of the first line 4 hours and 48 minutes $=\frac{24}{5}$ hours faster than the first line, then

$$
x-\frac{24}{5}=1:\left(\frac{1}{x-2}+\frac{1}{y}\right)
$$

After simplifications, this equation reduces to the following form:

$$
1+\frac{x-2}{y}=\frac{5 x-10}{5 x-24}
$$

We obtain the system of equations:

$$
\left\{\begin{array}{l}
\frac{1}{x}+\frac{1}{x-2}+\frac{1}{y}=1.5\left(\frac{1}{x}+\frac{1}{x-2}\right) \\
1+\frac{x-2}{y}=\frac{5 x-10}{5 x-24}
\end{array}\right.
$$

From the first equation of the system, express $y$ in terms of $x$:

$$
\frac{1}{y}=0.5\left(\frac{1}{x}+\frac{1}{x-2}\right), \quad \frac{1}{y}=\frac{1}{2} \cdot \frac{2 x-2}{x(x-2)}, \quad y=\frac{x(x-2)}{x-1}
$$

Substitute this expression into the second equation of the system:

$$
1+(x-2) \cdot \frac{x-1}{x(x-2)}=\frac{5 x-10}{5 x-24}, \quad 1+\frac{x-1}{x}=\frac{5 x-10}{5 x-24}
$$

The last equation reduces to the quadratic equation $5 x^{2}-43 x+24=0$. Its roots are $-x_{1}=8$ and $x_{2}=\frac{3}{5}$. Only the first root fits.

Answer: 8 hours.