Resistors in series and parallel conundrum
Start by assuming 2 of the 3 resistors have the same value (\$R_1\$) and so the series equation is: -

$$24 = 2R_1+R_2$$
And therefore

$$R_1 = \dfrac{24-R_2}{2}$$

Next substitute 2 lots of R1 in the parallel resistor equation: -

$$\dfrac{2}{24-R_2} +\dfrac{2}{24-R_2} + \dfrac{1}{R_2} = \dfrac{1}{1.8}$$

$$=\dfrac{4}{24-R_2} + \dfrac{1}{R_2} = \dfrac{1}{1.8}$$

$$=\dfrac{24-R_2+4R_2}{24R_2 - R_2^2} = \dfrac{1}{1.8}$$

Then manipulate the algebra and solve the implied quadratic equation to find a value of R2 of 2.7205 ohms (1st solution). This implies R1 = 10.63975 ohms.

In series (R1 + R1 + R2) they equal 24 ohms and, in parallel they equal 1.8000 ohms.

The 2nd solution to the quadratic yields R2 = 15.8795 ohms and this means R1 is 4.06025 ohms. In series they are 24 ohms and in parallel they are also 1.8 ohms.