If the points $P \left(6,2\right)$, $Q \left(-2,1\right)$ and $R$ are the vertices of $\triangle PQR$ and $R$ is the point on the locus $y = x^2 - 3x + 4$, then find the equation of locus of centroid of $\triangle PQR$.
Let $R \left(p,q\right)$ be the point on the locus.
$R$ satisfies the equation $y = x^2 - 3x + 4$
$\therefore$ $\;$ We have $\hspace{1em}$ $q = p^2 - 3p + 4$ $\;\;\; \cdots \; (1)$
Let $C \left(h,k\right)$ be a point on the equation of the locus of centroid of $\triangle PQR$.
Given: $P = \left(6,2\right)$ and $Q = \left(-2,1\right)$
$\therefore$ $\;$ Centroid of $\triangle PQR$ is
$\left(\dfrac{6 - 2 + p}{3}, \dfrac{2 + 1 + q}{3}\right) = \left(h, k\right)$
$\implies$ $\dfrac{4 + p}{3} = h$ $\implies$ $p = 3 h - 4$ $\;\;\; \cdots \; (2)$
and $\dfrac{3 + q}{3} = k$ $\implies$ $q = 3 k - 3$ $\;\;\; \cdots \; (3)$
Substituting the values of $p$ and $q$ from equations $(2)$ and $(3)$ in equation $(1)$ gives
$3 k - 3 = \left(3 h - 4\right)^2 - 3 \left(3 h - 4\right) + 4$
i.e. $3 k - 3 = 9 h^2 - 24 h + 16 - 9 h + 12 + 4$
i.e. $9 h^2 - 33 h - 3 k + 35 = 0$
$\therefore$ $\;$ The equation of locus of centroid of $\triangle PQR$ is: $\hspace{1em}$ $9x^2 - 33x - 3y + 35 = 0$