If the mean of the data 

is 21, then k is one of the roots of the equation :

The class midpoints are $$7.5, 12.5, 17.5, 22.5, 27.5, 32.5$$ respectively.

The total frequency is $$\sum f_i = 2 + k + 28 + 54 + (k+1) + 5 = 90 + 2k$$.

The sum $$\sum f_i x_i$$ is:

$$2(7.5) + k(12.5) + 28(17.5) + 54(22.5) + (k+1)(27.5) + 5(32.5)$$

$$= 15 + 12.5k + 490 + 1215 + 27.5k + 27.5 + 162.5 = 1910 + 40k$$

Setting the mean equal to 21:

$$\frac{1910 + 40k}{90 + 2k} = 21$$

$$1910 + 40k = 21(90 + 2k) = 1890 + 42k$$

$$20 = 2k$$

$$k = 10$$

Now we check which equation has $$x = 10$$ as a root. For Option 3: $$2x^2 - 19x - 10$$:

$$2(100) - 19(10) - 10 = 200 - 190 - 10 = 0$$

So $$k = 10$$ is indeed a root of $$2x^2 - 19x - 10 = 0$$.

Hence, the correct answer is Option 3.