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$\\Solution:\\ \text{Given: A circle inscribed in a triangle PQR in which PQ =12cm, QR =10cm and PR= 8cm}\\\\ \text{We know tangents drawn from an external point to a circle are equal in length.}\\ PL=PN\quad QL=QM\quad and\quad RM=RN\\\\ From\ figure:\\ PL+LQ=PQ\\ \Rightarrow PL+LQ=12cm\ \ \rightarrow (1)\\\\ RN+PN=PR\\ \Rightarrow RN+PL=8cm\ \ \rightarrow (2)\\\\ QM+RM=QR\\ \Rightarrow LQ+RN=10\ \ \rightarrow (3)\\\\ \text{Subtracting equation (2) from (1)}\\ \Rightarrow (LQ+RN)-( PL+LQ)=12-8\\ \Rightarrow LQ-RN=4\ \ \rightarrow (4)\\\\ \text{Adding equation (3) and (4)}\\\\ \Rightarrow (LQ+RN)+(LQ-RN)=10+4\\ \Rightarrow 2LQ=14\\ \therefore LQ=7cm\\\\ \text{Put LQ=7cm in equation (1) }\\ \Rightarrow PL+7=12\\ \Rightarrow PL=12-7\\ \therefore PL=5\\\\ \text{Put LQ=7cm in equation (3) }\\ \Rightarrow7+RN=10\\ \Rightarrow RN=10-7\\ \therefore RN=3\\\\ \therefore PL=PN=5cm\quad QL=QM=7cm\quad and\quad RM=RN=3cm$

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