Let $$L_1$$ be a tangent to the parabola $$y^2 = 4(x+1)$$ and $$L_2$$ be a tangent to the parabola $$y^2 = 8(x+2)$$ such that $$L_1$$ and $$L_2$$ intersect at right angles. Then $$L_1$$ and $$L_2$$ meet on the straight line:

We need two tangents, one to each parabola, which finally cut each other at a right angle. The first parabola is $$y^{2}=4(x+1)$$. For the standard parabola $$y^{2}=4ax$$ the tangent having slope $$m$$ is given by the well-known slope form

$$y=mx+\dfrac{a}{m}\,.$$

Here, after shifting the origin, $$a=1$$ and $$x$$ is replaced by $$x+1$$. So the tangent to the first parabola with slope $$m$$ is

$$y=m(x+1)+\dfrac{1}{m}\;. \quad -(1)$$

The second parabola is $$y^{2}=8(x+2)=4\cdot2\,(x+2)$$, hence $$a=2$$. Again using the slope form with the horizontal shift $$x\mapsto x+2$$, the tangent having slope $$n$$ is

$$y=n(x+2)+\dfrac{2}{n}\;. \quad -(2)$$

The two tangents meet at right angles, therefore the product of their slopes equals $$-1$$, i.e.

$$mn=-1\quad\Longrightarrow\quad n=-\dfrac1m\;. \quad -(3)$$

Let the two tangents intersect at the point $$P(h,k)$$. By definition, $$P$$ must satisfy both (1) and (2).

Substituting $$x=h,\;y=k$$ in (1) we get

$$k=m(h+1)+\dfrac1m\;. \quad -(4)$$

Likewise, substituting $$x=h,\;y=k$$ in (2) gives

$$k=n(h+2)+\dfrac{2}{n}\;. \quad -(5)$$

Because both right-hand sides equal $$k$$, we equate them:

$$m(h+1)+\dfrac1m=n(h+2)+\dfrac{2}{n}\;. \quad -(6)$$

Now we use relation (3), namely $$n=-1/m$$, on the right-hand side of (6).

Replacing $$n$$ by $$-1/m$$ gives

$$m(h+1)+\dfrac1m=-\dfrac{h+2}{m}-2m\;. \quad -(7)$$

To clear the denominators we multiply every term in (7) by $$m$$:

$$m^{2}(h+1)+1=-(h+2)-2m^{2}\;. \quad -(8)$$

Now we bring every term to the left side:

$$m^{2}(h+1)+1+(h+2)+2m^{2}=0\;. \quad -(9)$$

Combining the $$m^{2}$$ terms first, we notice

$$m^{2}(h+1)+2m^{2}=m^{2}\bigl((h+1)+2\bigr)=m^{2}(h+3)\;.$$

Similarly, collecting the constant terms, we have

$$1+(h+2)=h+3\;.$$

Hence equation (9) simplifies neatly to

$$(h+3)\bigl(m^{2}+1\bigr)=0\;. \quad -(10)$$

Because $$m^{2}+1$$ is always positive (it can never vanish), the only possibility for (10) to hold is

$$h+3=0\quad\Longrightarrow\quad h=-3\;. \quad -(11)$$

Thus the $$x$$-coordinate of the intersection point $$P$$ is fixed at $$-3$$, whereas the $$y$$-coordinate $$k$$ remains unrestricted. Consequently, every such pair of perpendicular tangents meets somewhere on the vertical line

$$x=-3\quad\Longleftrightarrow\quad x+3=0\;.$$

Hence, the correct answer is Option A.