$$\overline{r\ }=\left(3,4,1\right)+t\left(0,2,-1\right)$$ and $$P=\left(-4,7,6\right)$$ determine a plane $$\pi$$. What is the perpendicular distance from the point $$Q(1,1,1)$$ to $$\pi$$?

To find the perpendicular distance from the point $$Q(1, 1, 1)$$ to the plane $$\pi$$, we first need to determine the Cartesian equation of the plane.

The given equation of the line is:

$$\bar{r} = (3, 4, 1) + t(0, 2, -1)$$

From this equation, we can identify a fixed point on the line, let's call it $$A(3, 4, 1)$$, and the direction vector of the line:

$$\vec{d} = (0, 2, -1)$$

We are also given a point $$P(-4, 7, 6)$$ that lies on the plane $$\pi$$.

We can form a vector $$\vec{AP}$$ by joining point $$A$$ to point $$P$$:

$$\vec{AP} = P - A = (-4 - 3, 7 - 4, 6 - 1) = (-7, 3, 5)$$

The normal vector $$\vec{n}$$ to the plane $$\pi$$ is perpendicular to both the direction vector of the line $$\vec{d}$$ and the vector $$\vec{AP}$$.

We can find $$\vec{n}$$ by taking the cross product of these two vectors:

$$\vec{n} = \vec{d} \times \vec{AP} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 2 & -1 \\ -7 & 3 & 5 \end{vmatrix}$$

Expanding the determinant yields:

$$\vec{n} = \hat{i}(2(5) - (-1)(3)) - \hat{j}(0(5) - (-1)(-7)) + \hat{k}(0(3) - 2(-7))$$

$$\vec{n} = \hat{i}(10 + 3) - \hat{j}(0 - 7) + \hat{k}(0 + 14) = (13, 7, 14)$$

Using the normal vector $$\vec{n} = (13, 7, 14)$$ and the point $$A(3, 4, 1)$$, the equation of the plane $$\pi$$ is given by:

$$13(x - 3) + 7(y - 4) + 14(z - 1) = 0$$

$$13x - 39 + 7y - 28 + 14z - 14 = 0$$

$$13x + 7y + 14z - 81 = 0$$

Now, we calculate the perpendicular distance from the point $$Q(1, 1, 1)$$ to this plane using the standard distance formula:

$$D = \frac{\vert{}ax_1 + by_1 + cz_1 + d\vert{}}{\sqrt{a^2 + b^2 + c^2}}$$

Substituting the coordinates of $Q(1, 1, 1)$ and the coefficients of the plane into the formula:

$$D = \frac{\vert{}13(1) + 7(1) + 14(1) - 81\vert{}}{\sqrt{13^2 + 7^2 + 14^2}}$$

$$D = \frac{\vert{}13 + 7 + 14 - 81\vert{}}{\sqrt{169 + 49 + 196}}$$

$$D = \frac{\vert{}-47\vert{}}{\sqrt{414}} = \frac{47}{\sqrt{414}}$$

The correct option is A.