The incentre of the triangle whose vertices are \(\mathrm{P}(0,3,0), \mathrm{Q}(0,0,4)\) and \(\mathrm{R}(0,3,4)\) is

\(\begin{array}{ll} & \text { Let } \overline{\mathrm{p}}=3 \hat{\mathrm{j}}, \overline{\mathrm{q}}=4 \hat{\mathrm{k}}, \overline{\mathrm{r}}=3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}} \\ \therefore \quad & \overline{\mathrm{PQ}}=-3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}} \\ & \overline{\mathrm{QR}}=3 \hat{\mathrm{j}} \\ & \overline{\mathrm{PR}}=4 \hat{\mathrm{k}} \\ & \Rightarrow|\overline{\mathrm{PQ}}|=5,|\overline{\mathrm{QR}}|=3,|\overline{\mathrm{PR}}|=4\end{array}\)
Incentre of \(\triangle \mathrm{PQR}\) is given by
\(\begin{aligned}
& \frac{|\overline{\mathrm{PQ}}| \overrightarrow{\mathrm{r}}+|\overline{\mathrm{QR}}| \overline{\mathrm{p}}+|\overline{\mathrm{PR}}| \overrightarrow{\mathrm{q}}}{|\overline{\mathrm{PQ}}|+|\overline{\mathrm{QR}}|+|\overline{\mathrm{PR}}|} \\
& =\frac{5(3 \hat{\mathrm{j}}+4 \hat{\mathrm{k}})+3(3 \hat{\mathrm{j}})+4(4 \hat{\mathrm{k}})}{5+3+4} \\
& =\frac{24 \hat{\mathrm{j}}+36 \hat{\mathrm{k}}}{12} \\
& =2 \hat{\mathrm{j}}+3 \hat{\mathrm{k}} \\
\therefore \quad & \text { Incentre } \equiv(0,2,3)
\end{aligned}\)