A milkman has $21\text{litres}$ of whole milk, $42\text{litres}$ of toned milk and $63\text{litres}$ of double toned milk. If he wants to pack the three types of milk in cans so that in no can two types of milk are mixed, then what is the minimum number of cans he would require?

To find the minimum number of cans, the capacity of each can must be as large as possible. This maximum capacity equals the Highest Common Factor ($\text{HCF}$) of the quantities: $21$, $42$, and $63$.
$$\text{HCF}(21,42,63)=21\text{litres}$$

So, each can will hold $21\text{litres}$ of milk.
The number of cans required for each type of milk is:

$$\text{Cans for whole milk}=\frac{21}{21}=1$$

$$\text{Cans for toned milk}=\frac{42}{21}=2$$

$$\text{Cans for double toned milk}=\frac{63}{21}=3$$

$$\text{Total minimum cans required}=1+2+3=6$$