Which two numbers (not digits) should be interchanged to make the following equation correct?

$$21 \times 3 − 10 + 4 \div 7 = 5$$

$$21\times\ 3\ -\ 10\ +\ 4\ \frac{\ 7}{ }=5$$

21 and 4
$$21\times\ 3\ -\ 10\ +\ 4\ \frac{\ 7}{ }=5$$

$$4\times\ 3\ -\ 10\ +\ 21\ \frac{\ 7}{ }=5$$

$$4\times\ 3\ -\ 10\ +3=5$$

$$12-\ 10\ +3=5$$

$$2\ +3=5$$

$$5=5$$

4 and 7

$$21\times\ 3\ -\ 10\ +\ 4\ \frac{\ 7}{ }=5$$

$$21\times\ 3\ -\ 10\ +\ 7\ \frac{\ 4}{ }=5$$

$$63\ -\ 10\ +\ 7\ \frac{\ 4}{ }=5$$

$$53\ +\ 7\ \frac{\ 4}{ }=5$$

$$60\ \frac{\ 4}{ }=5$$

$$15=5$$

21 and 3

$$21\times\ 3\ -\ 10\ +\ 4\ \frac{\ 7}{ }=5$$

$$3\times\ 21\ -\ 10\ +\ 4\ \frac{\ 7}{ }=5$$

$$3\times11\ +\ 4\ \frac{\ 7}{ }=5$$

$$33+\ 4\ \frac{\ 7}{ }=5$$

$$\frac{37}{7}=5$$

$$5.285=5$$

3 and 7

$$21\times\ 3\ -\ 10\ +\ \frac{4}{7}=5$$

$$21\times\ 7-\ 10\ +\ \frac{4}{3}=5$$

$$147-\ 10\ +\ \frac{4}{3}=5$$

$$137\ +\ \frac{4}{3}=5$$

$$\frac{141}{3}=5$$

$$47=5$$

Hence option A is the correct answer