The equations of two sides of a variable triangle are $$x = 0$$ and $$y = 3$$, and its third side is a tangent to the parabola $$y^2 = 6x$$. The locus of its circumcentre is:

The two sides of the triangle are $$x = 0$$ (y-axis) and $$y = 3$$, and the third side is a tangent to the parabola $$y^2 = 6x$$.

Parametric tangent to $$y^2 = 6x$$.

For parabola $$y^2 = 4ax$$ where $$4a = 6$$, so $$a = \frac{3}{2}$$.

The tangent at parameter $$t$$ is:

$$ ty = x + \frac{3}{2}t^2 $$

Or equivalently: $$x = ty - \frac{3}{2}t^2$$

Find the vertices of the triangle.

Vertex A (intersection of $$x = 0$$ and $$y = 3$$): $$A = (0, 3)$$

Vertex B (intersection of tangent with $$x = 0$$):

$$ 0 = ty - \frac{3}{2}t^2 \implies y = \frac{3t}{2} $$

So $$B = \left(0, \frac{3t}{2}\right)$$

Vertex C (intersection of tangent with $$y = 3$$):

$$ x = 3t - \frac{3}{2}t^2 $$

So $$C = \left(3t - \frac{3}{2}t^2, 3\right)$$

Find the circumcenter.

The triangle has a right angle at $$A = (0, 3)$$ since the two sides $$x = 0$$ and $$y = 3$$ are perpendicular.

For a right triangle, the circumcenter is the midpoint of the hypotenuse (the side opposite the right angle, which is the tangent line segment $$BC$$).

Circumcenter $$= \left(\frac{0 + 3t - \frac{3}{2}t^2}{2}, \frac{\frac{3t}{2} + 3}{2}\right)$$

$$ h = \frac{3t - \frac{3}{2}t^2}{2} = \frac{3t(2 - t)}{4} $$ $$ k = \frac{\frac{3t}{2} + 3}{2} = \frac{3t + 6}{4} = \frac{3(t + 2)}{4} $$

Eliminate parameter $$t$$.

From $$k = \frac{3(t+2)}{4}$$:

$$ 4k = 3t + 6 $$ $$ t = \frac{4k - 6}{3} $$

Substituting into the expression for $$h$$:

$$ h = \frac{3t(2 - t)}{4} $$

Let me compute $$t(2 - t)$$:

$$ t(2 - t) = \frac{(4k-6)}{3} \cdot \left(2 - \frac{4k-6}{3}\right) = \frac{(4k-6)}{3} \cdot \frac{6 - 4k + 6}{3} = \frac{(4k-6)(12 - 4k)}{9} $$ $$ = \frac{-4(4k-6)(k-3)}{9} \cdot \text{... let me redo} $$

$$2 - t = 2 - \frac{4k-6}{3} = \frac{6 - 4k + 6}{3} = \frac{12 - 4k}{3}$$

$$ t(2-t) = \frac{(4k-6)(12-4k)}{9} $$ $$ h = \frac{3}{4} \cdot \frac{(4k-6)(12-4k)}{9} = \frac{(4k-6)(12-4k)}{12} $$

Expanding the numerator:

$$ (4k-6)(12-4k) = 48k - 16k^2 - 72 + 24k = -16k^2 + 72k - 72 $$ $$ h = \frac{-16k^2 + 72k - 72}{12} $$ $$ 12h = -16k^2 + 72k - 72 $$

Replacing $$(h, k)$$ with $$(x, y)$$:

$$ 12x = -16y^2 + 72y - 72 $$ $$ 16y^2 - 72y + 12x + 72 = 0 $$ $$ 4(4y^2 - 18y + 3x + 18) = 0 $$ $$ 4y^2 - 18y + 3x + 18 = 0 $$

This matches Option 3: $$4y^2 - 18y + 3x + 18 = 0$$.

The answer is $$\boxed{4y^2 - 18y + 3x + 18 = 0}$$.