The maximum error in the measurement of resistance, current and time for which current flows in an electrical circuit are $$1\%$$, $$2\%$$ and $$3\%$$ respectively. The maximum percentage error in the detection of the dissipated heat will be:

The heat dissipated in a resistor is given by:

$$H = I^2 R t$$

Taking the logarithm of both sides and differentiating yields the formula for the percentage error in H:

$$\frac{\Delta H}{H} \times 100 = 2\left(\frac{\Delta I}{I} \times 100\right) + \frac{\Delta R}{R} \times 100 + \frac{\Delta t}{t} \times 100$$

Since $$\frac{\Delta R}{R} \times 100 = 1\%$$, $$\frac{\Delta I}{I} \times 100 = 2\%$$, and $$\frac{\Delta t}{t} \times 100 = 3\%$$, substituting these values gives:

$$\frac{\Delta H}{H} \times 100 = 2(2\%) + 1\% + 3\% = 4\% + 1\% + 3\% = 8\%$$

Therefore, the maximum percentage error in the dissipated heat is $$8\%$$, which corresponds to Option D.