The mean of 5 observations is 5 and their variance is 12.4. If three of the observations are 1, 2 & 6; then the value of the remaining two is:

We are told that there are five observations whose mean is $$5$$. By definition,

$$\text{Mean}=\dfrac{\text{Sum of all observations}}{\text{Number of observations}}.$$

So, if we let the five observations be $$1,\,2,\,6,\,x,\,y,$$ then

$$\dfrac{1+2+6+x+y}{5}=5.$$

Multiplying both sides by $$5$$ gives

$$1+2+6+x+y=25.$$

Simplifying the left‐hand side, we obtain

$$9+x+y=25,$$

so that

$$x+y=25-9=16.$$

Thus the sum of the two unknown observations is $$16.$$

Next, we use the information about the variance. For a finite set of $$n$$ observations, the variance is defined as

$$\sigma^{2}=\dfrac{\sum\limits_{i=1}^{n}(x_i-\bar x)^2}{n},$$

where $$\bar x$$ is the mean. Here $$n=5$$ and $$\sigma^{2}=12.4,$$ so

$$\sum_{i=1}^{5}(x_i-5)^2 = n\sigma^{2}=5\times 12.4 = 62.$$

We already know three of the deviations from the mean:

$$\begin{aligned} (1-5)^2 &= (-4)^2 = 16,\\ (2-5)^2 &= (-3)^2 = 9,\\ (6-5)^2 &= 1^2 = 1. \end{aligned}$$

Adding these, the contribution of the known observations to the total is

$$16+9+1=26.$$

Therefore, the remaining two observations must contribute

$$62-26 = 36$$

to the total sum of squared deviations. In symbols,

$$(x-5)^2 + (y-5)^2 = 36.$$

We now have two equations:

$$\begin{cases} x+y = 16,\\ (x-5)^2 + (y-5)^2 = 36. \end{cases}$$

To solve, first expand the second equation:

$$x^2 - 10x + 25 + y^2 - 10y + 25 = 36.$$

Combine like terms:

$$x^2 + y^2 - 10x - 10y + 50 = 36.$$

Since $$x+y=16,$$ replace $$-10x-10y$$ by $$-10(x+y)=-10(16)=-160.$$ Substituting yields

$$x^2 + y^2 - 160 + 50 = 36,$$

which simplifies to

$$x^2 + y^2 - 110 = 36,$$

and therefore

$$x^2 + y^2 = 146.$$

Now square the first equation $$x+y=16$$ to obtain

$$(x+y)^2 = 16^2 = 256.$$

But we also know that

$$(x+y)^2 = x^2 + y^2 + 2xy.$$

Substituting $$x^2+y^2=146$$ into this identity gives

$$256 = 146 + 2xy,$$

so

$$2xy = 256 - 146 = 110,$$

and hence

$$xy = 55.$$

Therefore, $$x$$ and $$y$$ satisfy the quadratic equation whose roots have sum $$16$$ and product $$55$$:

$$t^2 - 16t + 55 = 0.$$

Compute the discriminant:

$$\Delta = 16^2 - 4\cdot 55 = 256 - 220 = 36.$$

Since $$\sqrt\Delta = 6,$$ the roots are

$$t = \dfrac{16 \pm 6}{2}.$$

This gives

$$t = \dfrac{22}{2}=11 \quad\text{or}\quad t = \dfrac{10}{2}=5.$$

Hence the two missing observations are $$5$$ and $$11.$$

Comparing with the choices, these correspond to Option C $$\bigl(5,\,11\bigr).$$

Hence, the correct answer is Option C.