Any parametric point on the line $$L$$ can be expressed by equating the symmetric equations to a scalar parameter $k$:
$$x = 2k + 1, \quad y = -k + 2, \quad z = k + 3$$
Since point $$A$$ lies on the plane $$P$$, substitute these coordinates into the plane equation $$x + y + z = 6$$:
$$(2k + 1) + (-k + 2) + (k + 3) = 6$$ $$2k + 6 = 6 \implies k = 0$$
Substituting $$k = 0$$ back into our parametric expressions gives the coordinates of point $$A$$:
$$A = (1, 2, 3)$$
The parabola lies in the $$xy$$-plane ($$z = 0$$) and has the equation $$y^2 = 4x$$. We can express any point $$B$$ on this parabola using its standard parametric form $$(t^2, 2t)$$:
$$B = (t^2, 2t, 0)$$
Using the 3D distance formula, write the squared distance $$D^2$$ between $$A(1, 2, 3)$$ and $$B(t^2, 2t, 0)$$:
$$D^2 = (t^2 - 1)^2 + (2t - 2)^2 + (0 - 3)^2$$ $$D^2 = (t^4 - 2t^2 + 1) + (4t^2 - 8t + 4) + 9$$ $$D^2 = t^4 + 2t^2 - 8t + 14$$
To find the minimum value, differentiate $$D^2$$ with respect to $$t$$ and set it to $$0$$:
$$\frac{d(D^2)}{dt} = 4t^3 + 4t - 8 = 0$$ $$t^3 + t - 2 = 0$$
By inspection, $$t = 1$$ is a real root of this cubic equation because $$1^3 + 1 - 2 = 0$$. Factoring out $$(t-1)$$ yields:
$$(t - 1)(t^2 + t + 2) = 0$$
Since the discriminant of the quadratic factor $$t^2 + t + 2$$ is negative ($$1^2 - 4(1)(2) = -7 < 0$$), $$t = 1$$ is the only real critical point.
Substitute $$t = 1$$ back into our distance squared equation:
$$D^2 = (1)^4 + 2(1)^2 - 8(1) + 14 = 1 + 2 - 8 + 14 = 9$$ $$D = \sqrt{9} = 3$$
Correct Option: B
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