The point of intersection of the normals to the parabola $$y^2 = 4x$$ at the ends of its latus rectum is :

The given parabola is $$ y^2 = 4x $$. Comparing this with the standard form $$ y^2 = 4ax $$, we find that $$ 4a = 4 $$, so $$ a = 1 $$. The focus of the parabola is at $$ (a, 0) = (1, 0) $$. The latus rectum is the line segment perpendicular to the axis passing through the focus, so it is the line $$ x = a $$, which is $$ x = 1 $$.

To find the endpoints of the latus rectum, substitute $$ x = 1 $$ into the parabola equation: $$ y^2 = 4 \times 1 = 4 $$. Solving for $$ y $$, we get $$ y = \pm 2 $$. Thus, the endpoints are $$ (1, 2) $$ and $$ (1, -2) $$.

Next, we need the equations of the normals to the parabola at these points. Using the parametric form, a point on the parabola $$ y^2 = 4ax $$ can be written as $$ (at^2, 2at) $$. Since $$ a = 1 $$, the points are $$ (t^2, 2t) $$.

For the point $$ (1, 2) $$, we have $$ 2t = 2 $$, so $$ t = 1 $$. For the point $$ (1, -2) $$, we have $$ 2t = -2 $$, so $$ t = -1 $$.

The equation of the normal to the parabola $$ y^2 = 4ax $$ at parameter $$ t $$ is given by $$ y = tx - 2at - at^3 $$. Substituting $$ a = 1 $$, this simplifies to $$ y = tx - 2t - t^3 $$.

For $$ t = 1 $$: $$ y = (1)x - 2(1) - (1)^3 = x - 2 - 1 = x - 3 $$. So, the normal at $$ (1, 2) $$ is $$ y = x - 3 $$.

For $$ t = -1 $$: $$ y = (-1)x - 2(-1) - (-1)^3 = -x + 2 - (-1) = -x + 2 + 1 = -x + 3 $$. So, the normal at $$ (1, -2) $$ is $$ y = -x + 3 $$.

To find the point of intersection of these two normals, solve the system of equations:

Equation 1: $$ y = x - 3 $$

Equation 2: $$ y = -x + 3 $$

Set them equal: $$ x - 3 = -x + 3 $$. Add $$ x $$ to both sides: $$ x + x - 3 = 3 $$, which simplifies to $$ 2x - 3 = 3 $$. Add 3 to both sides: $$ 2x = 6 $$. Divide by 2: $$ x = 3 $$.

Substitute $$ x = 3 $$ into Equation 1: $$ y = 3 - 3 = 0 $$. Thus, the point of intersection is $$ (3, 0) $$.

Verifying with Equation 2: when $$ x = 3 $$, $$ y = -3 + 3 = 0 $$, which matches.

Comparing with the options:

A. $$ (0, 2) $$

B. $$ (3, 0) $$

C. $$ (0, 3) $$

D. $$ (2, 0) $$

Hence, the correct answer is Option B.