Three numbers are such that if the average of any two of them is added to the third number, the sums obtained are 164, 158 and 132 respectively. What is the average of the original three numbers?

Let the three numbers be $$x$$, $$y$$ and $$z$$.

The statements in the question translate to the following three equations:

Average of $$x$$ and $$y$$ added to $$z$$ gives 164:
$$\frac{x+y}{2}+z=164$$

Average of $$y$$ and $$z$$ added to $$x$$ gives 158:
$$\frac{y+z}{2}+x=158$$

Average of $$z$$ and $$x$$ added to $$y$$ gives 132:
$$\frac{z+x}{2}+y=132$$

Clear the fractions (multiply every equation by 2):

$$x+y+2z=328$$ $$-(1)$$

$$y+z+2x=316$$ $$-(2)$$

$$z+x+2y=264$$ $$-(3)$$

Add $$-(1),\; -(2)$$ and $$-(3)$$ to eliminate the individual variables:

Left side
$$\bigl(x+y+2z\bigr)+\bigl(y+z+2x\bigr)+\bigl(z+x+2y\bigr)$$
Coefficients collected give $$4x+4y+4z = 4(x+y+z)$$.

Right side
$$328+316+264 = 908$$.

Therefore,
$$4(x+y+z)=908 \quad\Longrightarrow\quad x+y+z=\frac{908}{4}=227$$

The average of the three original numbers is
$$\frac{x+y+z}{3}=\frac{227}{3}=75\frac{2}{3}$$

Hence the required average is $$75\frac{2}{3}$$.
Option C which is: $$75\frac{2}{3}$$