Solution:

According to the given question,

ABC is a triangle with \[AB\text{ }=\text{ }10\text{ }cm,\text{ }BC=\text{ }8\text{ }cm,\text{ }AC\text{ }=\text{ }6\text{ }cm\]

Three circles are drawn with centre \[A,\text{ }B\text{ }and\text{ }C\]touch each other at \[P,\text{ }Q\text{ }and\text{ }R\] respectively.

So, we need to find the radii of the three circles.

Let,

\[PA\text{ }=\text{ }AQ\text{ }=\text{ }x\]

\[QC\text{ }=\text{ }CR\text{ }=\text{ }y\]

\[RB\text{ }=\text{ }BP\text{ }=\text{ }z\]

So, we have

\[x\text{ }+\text{ }z\text{ }=\text{ }10\text{ }\ldots ..\text{ }\left( i \right)\]

\[z\text{ }+\text{ }y\text{ }=\text{ }8\text{ }\ldots \ldots \text{ }\left( ii \right)\]

\[y\text{ }+\text{ }x\text{ }=\text{ }6\text{ }\ldots \ldots .\text{ }\left( iii \right)\]

Adding all the three equations, we have

\[2\left( x\text{ }+\text{ }y\text{ }+\text{ }z \right)\text{ }=\text{ }24\]

\[x\text{ }+\text{ }y\text{ }+\text{ }z\text{ }=\text{ }24/2\text{ }=\text{ }12\text{ }\ldots ..\text{ }\left( iv \right)\]

Subtracting (i), (ii) and (iii) from (iv) we get

\[y\text{ }=\text{ }12\text{ }-\text{ }10\text{ }=\text{ }2\]

\[x\text{ }=\text{ }12\text{ }-\text{ }8\text{ }=\text{ }4\]

And,

\[z\text{ }=\text{ }12\text{ }-\text{ }6\text{ }=\text{ }6\]

Thus, radii of the three circles are \[2\text{ }cm,\text{ }4\text{ }cm\text{ }and\text{ }6\text{ }cm\]