Given points Q=(1, 4), R=(3, -2)

Let P(h, k) be third vertex which lies on the line $2\mathrm{x}-3\mathrm{y}+4=0$

$\Rightarrow 2\mathrm{h}-3\mathrm{k}+4=0 - \left(1\right)$

Let $\mathrm{G}({\mathrm{x}}_1, {\mathrm{y}}_1)$ be locus of centriod of ${∆}^{\mathrm{le}} \mathrm{PQR}$

$$\begin{gathered} \Rightarrow ({\mathrm{x}}_1, {\mathrm{y}}_1)=\left(\frac{1+3+\mathrm{h}}{3}, \frac{4-2+\mathrm{k}}{3}\right) \\ \Rightarrow {\mathrm{x}}_1=\frac{4+\mathrm{h}}{3}, {\mathrm{y}}_1=\frac{2+\mathrm{k}}{3} \\ \Rightarrow 3{\mathrm{x}}_1=4+\mathrm{h}, 3{\mathrm{y}}_1=2+\mathrm{k} \\ \Rightarrow \mathrm{h}=3{\mathrm{x}}_1-4, \mathrm{k}=3{\mathrm{y}}_1-2 \\ \mathrm{from} \left(1\right) 2(3{\mathrm{x}}_1-2)+4= \\ \Rightarrow 6{\mathrm{x}}_1-8-9{\mathrm{y}}_1+6+4=0 \\ \Rightarrow 6{\mathrm{x}}_1-9{\mathrm{y}}_1+2=0 \Rightarrow \mathrm{slope} =\frac{-6}{-9}=\frac{2}{3} \end{gathered}$$

 $\therefore$Locus of ‘G’ is a line with slope $\frac{2}{3}$