The tangents to the curve $$y = (x - 2)^2 - 1$$ at its points of intersection with the line $$x - y = 3$$, intersect at the point:

We begin with the curve $$y = (x - 2)^2 - 1$$ and the straight line $$x - y = 3$$. To locate their points of intersection, we substitute the expression for $$y$$ from the line into the curve.

The line gives $$y = x - 3$$. Substituting this in the curve’s equation, we obtain

$$x - 3 = (x - 2)^2 - 1.$$

Expanding the square on the right side,

$$(x - 2)^2 = x^2 - 4x + 4,$$

so the equation becomes

$$x - 3 = x^2 - 4x + 4 - 1.$$

Simplifying the right side, we have

$$x - 3 = x^2 - 4x + 3.$$

Now, bringing every term to the right,

$$0 = x^2 - 4x + 3 - x + 3 = x^2 - 5x + 6.$$

Factoring the quadratic,

$$x^2 - 5x + 6 = (x - 2)(x - 3) = 0.$$

This gives two real solutions:

$$x = 2 \quad \text{or} \quad x = 3.$$

Substituting back in $$y = x - 3:$$

If $$x = 2,$$ then $$y = 2 - 3 = -1,$$ giving the point $$P_1(2,\,-1).$$

If $$x = 3,$$ then $$y = 3 - 3 = 0,$$ giving the point $$P_2(3,\,0).$$

Next, we need the tangents to the curve at these points. For that, we first compute the derivative of the curve.

The curve is $$y = (x - 2)^2 - 1.$$ Using the power rule,

$$\frac{dy}{dx} = 2(x - 2).$$

At $$P_1(2,\,-1),$$ the slope is

$$m_1 = 2(2 - 2) = 2 \cdot 0 = 0.$$

Thus the tangent at $$P_1$$ is a horizontal line passing through $$y = -1$$, i.e.

$$y = -1.$$

At $$P_2(3,\,0),$$ the slope is

$$m_2 = 2(3 - 2) = 2 \cdot 1 = 2.$$

Using point-slope form $$y - y_0 = m(x - x_0),$$ we write the tangent at $$P_2$$ as

$$y - 0 = 2(x - 3) \quad\Longrightarrow\quad y = 2x - 6.$$

The final task is to find the intersection of these two tangents. One tangent is $$y = -1$$ and the other is $$y = 2x - 6.$$ Setting them equal, we get

$$-1 = 2x - 6.$$

Adding $$6$$ to both sides,

$$5 = 2x,$$

and dividing by $$2,$$

$$x = \frac{5}{2}.$$

Substituting this back into $$y = -1$$ (the simpler of the two tangents) immediately gives

$$y = -1.$$

Thus the tangents intersect at the point $$\left(\dfrac{5}{2},\, -1\right).$$

Hence, the correct answer is Option B.