1. Interpretation of the question

We must choose exactly $5$ fruits from:

• $7$ red apples,
• $5$ white apples,
• $8$ oranges,

subject to:

• at least $2$ oranges,
• at least $1$ red apple,
• at least $1$ white apple.

Since fruits of the same type/color are distinct individual fruits from the basket, we must count selections using combinations.

Let:

• $r$ = number of red apples chosen,
• $w$ = number of white apples chosen,
• $o$ = number of oranges chosen.

Then $r+w+o=5$ with constraints $o\ge 2,\quad r\ge 1,\quad w\ge 1.$

2. Find all possible distributions $(r,w,o)$

Because $r\ge 1$, $w\ge 1$, $o\ge 2$ and total is $5$:

First allocate the minimum required fruits: $1\text{ red} + 1\text{ white} + 2\text{ oranges} = 4$

So only $1$ fruit remains to be assigned.

Thus the only possible cases are:

1. $(r,w,o)=(2,1,2)$
2. $(r,w,o)=(1,2,2)$
3. $(r,w,o)=(1,1,3)$

3. Count each case

Case 1: $(2,1,2)$

Choose:

• $2$ red apples from $7$: $\binom{7}{2}$
• $1$ white apple from $5$: $\binom{5}{1}$
• $2$ oranges from $8$: $\binom{8}{2}$

Number of ways: $\binom{7}{2}\binom{5}{1}\binom{8}{2} = 21\cdot 5\cdot 28 = 2940$

Case 2: $(1,2,2)$

Choose:

• $1$ red apple from $7$: $\binom{7}{1}$
• $2$ white apples from $5$: $\binom{5}{2}$
• $2$ oranges from $8$: $\binom{8}{2}$

Number of ways: $\binom{7}{1}\binom{5}{2}\binom{8}{2} = 7\cdot 10\cdot 28 = 1960$

Case 3: $(1,1,3)$

Choose:

• $1$ red apple from $7$: $\binom{7}{1}$
• $1$ white apple from $5$: $\binom{5}{1}$
• $3$ oranges from $8$: $\binom{8}{3}$

Number of ways: $\binom{7}{1}\binom{5}{1}\binom{8}{3} = 7\cdot 5\cdot 56 = 1960$

4. Add all cases

Total number of ways: $2940+1960+1960=6860$

5. Compare with stored correct answer

Stored correct answer: $6860$ OR $3$

Our derived answer is: $\boxed{6860}$

So it agrees with the valid stored answer $6860$. The value $3$ appears to correspond only to the number of possible fruit-type distributions, not the number of actual selections of fruits.