If the vertices P and Q of a triangle PQR are given by (2, 5) and $\left(4,-11\right)$ , respectively, and the point R moves along the line N given by $9x+7y+4=0$ , then the locus of the centroid of the triangle PQR is a straight line parallel to

$R\left(x,y\right)$ lies on $9x+7y+4=0$

If x-cordinate is ‘a’ , $R\left(a, \frac{-(4+9a)}{7}\right),a\in R$

If (h,k) is the centroid 

$(h,k)=\left(\frac{2+4+a}{3}, \frac{5-11+{\displaystyle \frac{(-4-9a)}{7}}}{3}\right)$

$h=\frac{2+4+a}{3}=\frac{6+a}{3}$ ----------- (1)

$k=\frac{5-11-\frac{\left(4+9a\right)}{7}}{3}$$=\frac{-46-9a}{7\times 3}$ -------------(2)

From (1) and (2) eliminating 'a'  we get

$3h-6=\frac{-\left(21k-46\right)}{9}$

$\Rightarrow 27h+21k-54+46=0$

Hence, the locus is $9x+7y-\frac{8}{3}=0$ . This line is parallel to N.