If the standard deviation of the numbers 2, 3, $$a$$ and 11 is 3.5, then which of the following is true?

We have the four numbers $$2,\;3,\;a,\;11$$. Their arithmetic mean is obtained from the formula for the mean of $$n$$ observations, $$\mu=\dfrac{\text{sum of the observations}}{n}$$. Here $$n=4$$, so

$$ \mu=\dfrac{2+3+a+11}{4} =\dfrac{16+a}{4} =\dfrac{a+16}{4}. $$

The standard deviation $$\sigma$$ of a set of four (population) values is defined by

$$ \sigma=\sqrt{\dfrac {(x_1-\mu)^2+(x_2-\mu)^2+(x_3-\mu)^2+(x_4-\mu)^2}{4}}, $$

and its square, the variance, is therefore

$$ \sigma^2=\dfrac {(x_1-\mu)^2+(x_2-\mu)^2+(x_3-\mu)^2+(x_4-\mu)^2}{4}. $$

We are told that $$\sigma=3.5$$, so $$\sigma^2=(3.5)^2=12.25=\dfrac{49}{4}$$.

Now let us write each squared deviation.

First deviation:

$$ 2-\mu=2-\dfrac{a+16}{4}=\dfrac{8-a-16}{4}=-\dfrac{a+8}{4}, \qquad (2-\mu)^2=\dfrac{(a+8)^2}{16}. $$

Second deviation:

$$ 3-\mu=3-\dfrac{a+16}{4}=\dfrac{12-a-16}{4}=-\dfrac{a+4}{4}, \qquad (3-\mu)^2=\dfrac{(a+4)^2}{16}. $$

Third deviation:

$$ a-\mu=a-\dfrac{a+16}{4}=\dfrac{4a-a-16}{4}=\dfrac{3a-16}{4}, \qquad (a-\mu)^2=\dfrac{(3a-16)^2}{16}. $$

Fourth deviation:

$$ 11-\mu=11-\dfrac{a+16}{4}=\dfrac{44-a-16}{4}=\dfrac{28-a}{4}, \qquad (11-\mu)^2=\dfrac{(28-a)^2}{16}. $$

Adding the four squared deviations gives

$$ \begin{aligned} &(2-\mu)^2+(3-\mu)^2+(a-\mu)^2+(11-\mu)^2 \\[4pt] &=\dfrac{(a+8)^2}{16}+\dfrac{(a+4)^2}{16}+\dfrac{(3a-16)^2}{16}+\dfrac{(28-a)^2}{16} \\[4pt] &=\dfrac{(a+8)^2+(a+4)^2+(3a-16)^2+(28-a)^2}{16}. \end{aligned} $$

Placing this expression into the variance formula we get

$$ \sigma^2=\dfrac{1}{4}\times\dfrac{(a+8)^2+(a+4)^2+(3a-16)^2+(28-a)^2}{16} =\dfrac{(a+8)^2+(a+4)^2+(3a-16)^2+(28-a)^2}{64}. $$

Since $$\sigma^2=\dfrac{49}{4}$$, we equate the two expressions:

$$ \dfrac{(a+8)^2+(a+4)^2+(3a-16)^2+(28-a)^2}{64}=\dfrac{49}{4}. $$

Multiplying both sides by $$64$$ produces

$$ (a+8)^2+(a+4)^2+(3a-16)^2+(28-a)^2=64\times\dfrac{49}{4}=16\times49=784. $$

Now expand every square and collect like terms:

$$\begin{aligned} (a+8)^2 &= a^2+16a+64$$, $$\\ (a+4)^2 &= a^2 \;+\; 8a \;+\;16$$, $$\\ (3a-16)^2 &= 9a^2-96a+256$$, $$\\ (28-a)^2 &= a^2-56a+784. \end{aligned}$$

Adding them yields

$$\begin{aligned} &[a^2+16a+64]+\$$, $$[a^2+8a+16]+\$$, $$[9a^2-96a+256]+\$$, $$[a^2-56a+784] \\[4pt] &=(a^2+a^2+9a^2+a^2)+(16a+8a-96a-56a)+(64+16+256+784) \\[4pt] &=12a^2-128a+1120. \end{aligned}$$

Equating this sum to $$784$$ gives

$$ 12a^2-128a+1120=784. $$

Subtracting $$784$$ from both sides, we obtain

$$ 12a^2-128a+336=0. $$

Dividing every term by $$4$$ simplifies the quadratic to

$$ 3a^2-32a+84=0. $$

This relation matches Option D, which is written exactly as $$3a^2 - 32a + 84 = 0$$.

Hence, the correct answer is Option D.