If the foot of the perpendicular from the point $$A(-1, 4, 3)$$ on the plane $$P: 2x + my + nz = 4$$, is $$\left(-2, \frac{7}{2}, \frac{3}{2}\right)$$, then the distance of the point A from the plane P, measured parallel to a line with direction ratios 3, -1, -4, is equal to

We are given the point $$A(-1, 4, 3)$$, the plane $$P: 2x + my + nz = 4$$, and the foot of the perpendicular from $$A$$ to $$P$$ is $$F\left(-2, \frac{7}{2}, \frac{3}{2}\right)$$.

The direction of the perpendicular from $$A$$ to the plane is along the normal to the plane, which has direction ratios $$(2, m, n)$$. The vector $$\vec{AF} = F - A = (-2-(-1),\;\frac{7}{2}-4,\;\frac{3}{2}-3) = (-1,\;-\frac{1}{2},\;-\frac{3}{2})$$.

Since $$\vec{AF}$$ must be parallel to $$(2, m, n)$$, we have $$\frac{-1}{2} = \frac{-1/2}{m} = \frac{-3/2}{n}$$. From $$\frac{-1}{2} = \frac{-1/2}{m}$$, we get $$m = 1$$. From $$\frac{-1}{2} = \frac{-3/2}{n}$$, we get $$n = 3$$.

So the plane is $$2x + y + 3z = 4$$. We verify: the foot $$F$$ should lie on the plane: $$2(-2) + \frac{7}{2} + 3\cdot\frac{3}{2} = -4 + \frac{7}{2} + \frac{9}{2} = -4 + 8 = 4$$. Correct.

Now we need the distance from $$A(-1, 4, 3)$$ to the plane $$P$$, measured parallel to a line with direction ratios $$(3, -1, -4)$$. The parametric line through $$A$$ in this direction is $$(x, y, z) = (-1+3t,\;4-t,\;3-4t)$$.

We substitute into the plane equation: $$2(-1+3t) + (4-t) + 3(3-4t) = 4$$, which gives $$-2+6t+4-t+9-12t = 4$$, so $$11 - 7t = 4$$, hence $$t = 1$$.

The point of intersection is $$(-1+3,\;4-1,\;3-4) = (2, 3, -1)$$. The distance from $$A$$ to this point is:

$$d = \sqrt{(2-(-1))^2 + (3-4)^2 + (-1-3)^2} = \sqrt{9 + 1 + 16} = \sqrt{26}$$

Hence, the correct answer is Option B.