Find the locus of point R such that 5PR = 3QR, given P(2, 5) and Q(4, 7).

Let $R(x, y)$ be the variable point. The fixed points are $P(2, 5)$ and $Q(4, 7)$.
The distance formula between two points $(x_1, y_1)$ and $(x_2, y_2)$ is $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$.

For $|\vec{PR}|$:

$$|\vec{PR}| = \sqrt{(x-2)^2 + (y-5)^2}$$

For $|\vec{QR}|$:

$$|\vec{QR}| = \sqrt{(x-4)^2 + (y-7)^2}$$

$$\left(5\sqrt{(x-2)^2 + (y-5)^2}\right)^2 = \left(3\sqrt{(x-4)^2 + (y-7)^2}\right)^2$$ $$25\left((x-2)^2 + (y-5)^2\right) = 9\left((x-4)^2 + (y-7)^2\right)$$

$$(x-2)^2 = x^2 - 4x + 4$$ $$(y-5)^2 = y^2 - 10y + 25$$ $$(x-4)^2 = x^2 - 8x + 16$$ $$(y-7)^2 = y^2 - 14y + 49$$

Substitute these expansions back into the equation:

$$25(x^2 - 4x + 4 + y^2 - 10y + 25) = 9(x^2 - 8x + 16 + y^2 - 14y + 49)$$ $$25(x^2 - 4x + y^2 - 10y + 29) = 9(x^2 - 8x + y^2 - 14y + 65)$$

Subtract $9x^2$, $-72x$, $9y^2$, $-126y$, and 585 from both sides:

$$(25x^2 - 9x^2) + (-100x + 72x) + (25y^2 - 9y^2) + (-250y + 126y) + (725 - 585) = 0$$ $$16x^2 - 28x + 16y^2 - 124y + 140 = 0$$

$$\frac{16x^2}{4} - \frac{28x}{4} + \frac{16y^2}{4} - \frac{124y}{4} + \frac{140}{4} = 0$$ $$4x^2 - 7x + 4y^2 - 31y + 35 = 0$$

The equation of the locus of $R$ is:

$$\mathbf{4x^2 + 4y^2 - 7x - 31y + 35 = 0}$$