Let the mean and variance of 8 numbers - 10, - 7, - 1, x, y, 9, 2, 16 be $$\frac{7}{2}\text{ and }\frac{293}{4}$$ respectively.
Then the mean of 4 numbers x, y, x + y + 1, |x-y| is:

Sum of the eight numbers = $$-10 + (-7) + (-1) + x + y + 9 + 2 + 16 = 9 + x + y.$$

Mean is given by the formula $$\text{Mean} = \frac{\text{Sum of terms}}{n}$$. Here $$n=8$$ and the mean is $$\frac{7}{2}$$, so

$$\frac{9 + x + y}{8} = \frac{7}{2} \quad-(1)$$

Multiplying both sides of $$(1)$$ by 8 gives

$$9 + x + y = 28.$$

Therefore,

$$x + y = 28 - 9 = 19 \quad-(2)$$

Variance formula for $$n$$ numbers is

$$\sigma^2 = \frac{1}{n}\sum_{i=1}^n x_i^2 \;-\; (\text{mean})^2.$$

Here $$\sigma^2 = \frac{293}{4}$$, so

$$\frac{1}{8}\sum(\text{squares}) - \Bigl(\frac{7}{2}\Bigr)^2 = \frac{293}{4}\,.$$
Thus, $$\frac{1}{8}\sum(\text{squares}) = \frac{293}{4} + \frac{49}{4} = \frac{342}{4} = \frac{171}{2}\,.$$

Multiplying by 8,

$$\sum(\text{squares}) = 8 \times \frac{171}{2} = 684.$$

The sum of squares of the known constants is

$$(-10)^2 + (-7)^2 + (-1)^2 + 9^2 + 2^2 + 16^2 = 100 + 49 + 1 + 81 + 4 + 256 = 491.$$

Hence,

$$491 + x^2 + y^2 = 684\quad\Longrightarrow\quad x^2 + y^2 = 193 \quad-(3)$$

Using the identity $$(x + y)^2 = x^2 + y^2 + 2xy$$ and substituting $$(2)$$ and $$(3)$$:

$$19^2 = 193 + 2xy\;\Longrightarrow\;361 = 193 + 2xy\;\Longrightarrow\;2xy = 168\;\Longrightarrow\;xy = 84\quad-(4)$$

Now compute $$(x - y)^2$$ via

$$ (x - y)^2 = (x + y)^2 - 4xy = 361 - 4\times84 = 361 - 336 = 25.$$

Therefore,

$$|x - y| = 5.$$

The four numbers whose mean is required are $$x,\;y,\;x + y + 1,\;|x - y|$$. Their sum is

$$x + y + (x + y + 1) + |x - y| = 19 + 20 + 5 = 44.$$

Hence the mean of these four numbers is

$$\frac{44}{4} = 11.$$

Answer is Option A (11).