The average of three distinct real numbers is 28. If the smallest number is increased by 7 and the largest number is reduced by 10, the order of the numbers remains unchanged, and the new arithmetic mean becomes 2 more than the middle number, while the difference between the largest and the smallest numbers becomes 64. Then, the largest number in the original set of three numbers is:

The correct answer is 70.

Let the three distinct real numbers in ascending order be represented as $x$, $y$, and $z$, such that:
$x<y<z$

Here, $x$ is the smallest number, $y$ is the middle number, and $z$ is the largest number.

We are given that the average of these three numbers is 28. Therefore:
$\frac{x+y+z}{3}=28$
Multiplying both sides by 3, we get our first equation:
$x+y+z=84$   —  (Equation 1)

Next, the smallest number $x$ is increased by 7, becoming $x+7$.
The largest number $z$ is reduced by 10, becoming $z-10$.
The middle number remains unchanged as $y$.
Since the order of the numbers remains unchanged, the new smallest number is still the first term, the middle number is the second, and the new largest number is the third. Thus, the new set of numbers is:
$\{x+7,y,z-10\}$

Let us find the new arithmetic mean of these modified numbers:
New Mean = $\frac{(x+7)+y+(z-10)}{3}=\frac{x+y+z-3}{3}$
Using Equation 1, we can substitute $x+y+z=84$ into this expression:
New Mean = $\frac{84-3}{3}=\frac{81}{3}=27$

We are told that the new arithmetic mean is 2 more than the middle number $y$. Therefore:
$27=y+2$
Solving for $y$:
$y=25$

Now we substitute the value of $y=25$ back into Equation 1:
$x+25+z=84$
$x+z=59$   —  (Equation 2)

We are also given that the difference between the new largest number and the new smallest number is 64. Therefore:
$(z-10)-(x+7)=64$
Simplifying the expression:
$z-x-17=64$
$z-x=81$   —  (Equation 3)

Now, we have a system of two linear equations (Equation 2 and Equation 3) with two variables:
1) $z+x=59$
2) $z-x=81$

To find the largest number $z$ in the original set, we add the two equations together:
$(z+x)+(z-x)=59+81$
$2z=140$
$z=70$

Thus, the largest number in the original set of three numbers is 70.