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(C) The given pair of straight lines is $12x^2 - 20xy + 7y^2 = 0$. Factoring the quadratic expression: $12x^2 - 6xy - 14xy + 7y^2 = 0 \implies 6x(2x -

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$\left(\frac{8}{3}, \frac{8}{3}\right)$

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(C) The given pair of straight lines is $12x^2 - 20xy + 7y^2 = 0$. Factoring the quadratic expression: $12x^2 - 6xy - 14xy + 7y^2 = 0 \implies 6x(2x - y) - 7y(2x - y) = 0 \implies (6x - 7y)(2x - y) = 0$. Thus,the equations of the two sides are $L_1: 6x - 7y = 0$ and $L_2: 2x - y = 0$. The third side is $L_3: 2x - 3y + 4 = 0$. Solving $L_1$ and $L_2$: $6x - 7y = 0$ and $2x - y = 0 \implies x=0, y=0$. Vertex $A = (0, 0)$. Solving $L_2$ and $L_3$: $2x - y = 0 \implies y = 2x$. Substituting into $L_3$: $2x - 3(2x) + 4 = 0 \implies -4x = -4 \implies x = 1, y = 2$. Vertex $B = (1, 2)$. Solving $L_1$ and $L_3$: $6x - 7y = 0 \implies x = \frac{7y}{6}$. Substituting into $L_3$: $2(\frac{7y}{6}) - 3y + 4 = 0 \implies \frac{7y}{3} - 3y = -4 \implies -\frac{2y}{3} = -4 \implies y = 6, x = 7$. Vertex $C = (7, 6)$. The centroid $G$ is given by $\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right) = \left(\frac{0+1+7}{3}, \frac{0+2+6}{3}\right) = \left(\frac{8}{3}, \frac{8}{3}\right)$.

The centroid of the triangle formed by the pair of straight lines $12x^2 - 20xy + 7y^2 = 0$ and the line $2x - 3y + 4 = 0$ is:

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