The locus of the point of intersection of the straight lines, $$tx - 2y - 3t = 0$$ and $$x - 2ty + 3 = 0$$ ($$t \in R$$), is:

We are given two straight lines, each involving a real parameter $$t$$:

$$t\,x - 2y - 3t = 0 \qquad\text{and}\qquad x - 2t\,y + 3 = 0.$$

The required locus is the set of all intersection points $$(x,y)$$ obtained when $$t$$ takes every real value. To find this locus we must eliminate $$t$$ from the two equations.

First, from the equation $$t\,x - 2y - 3t = 0$$ we take all terms containing $$t$$ to one side:

$$t\,x - 3t = 2y.$$

Now factor out $$t$$ on the left:

$$t\,(x - 3) = 2y.$$

Provided $$x \neq 3$$, we can solve for $$t$$:

$$t = \dfrac{2y}{x - 3}.$$

Next, substitute this value of $$t$$ into the other straight-line equation $$x - 2t\,y + 3 = 0$$. We obtain:

$$x - 2\left(\dfrac{2y}{x - 3}\right)y + 3 = 0.$$

Simplifying the fraction inside the parentheses gives $$2 \cdot \dfrac{2y}{x - 3}\,y = \dfrac{4y^{2}}{x - 3}$$, so the expression becomes

$$x - \dfrac{4y^{2}}{x - 3} + 3 = 0.$$

To clear the denominator, multiply every term by $$x - 3$$:

$$(x - 3)\,x - 4y^{2} + (x - 3)\,3 = 0.$$

We now expand each product:

$$x^{2} - 3x - 4y^{2} + 3x - 9 = 0.$$

The terms $$-3x$$ and $$+3x$$ cancel, leaving

$$x^{2} - 4y^{2} - 9 = 0.$$

Rearrange to get all constants on the right:

$$x^{2} - 4y^{2} = 9.$$

This is the standard form of a rectangular conic. To see which one, we divide both sides by $$9$$:

$$\dfrac{x^{2}}{9} - \dfrac{4y^{2}}{9} = 1.$$

Write the second fraction as a single square in the denominator:

$$\dfrac{x^{2}}{9} - \dfrac{y^{2}}{\,9/4\,} = 1.$$

In the canonical form $$\dfrac{x^{2}}{a^{2}} - \dfrac{y^{2}}{b^{2}} = 1$$ for a hyperbola, we recognize

$$a^{2} = 9 \;\Longrightarrow\; a = 3, \qquad b^{2} = \dfrac{9}{4} \;\Longrightarrow\; b = \dfrac{3}{2}.$$

For a hyperbola, the conjugate axis has length $$2b$$. Therefore

$$2b = 2\left(\dfrac{3}{2}\right) = 3.$$

Thus the locus is a hyperbola whose conjugate axis is exactly $$3$$ units long.

Hence, the correct answer is Option A.