The square of the distance of the point of intersection of the line $$\frac{x-1}{2} = \frac{y-2}{3} = \frac{z+1}{6}$$ and the plane $$2x - y + z = 6$$ from the point $$(-1, -1, 2)$$ is _________.

We begin with the symmetric (or proportional) form of the line

$$\frac{x-1}{2} = \frac{y-2}{3} = \frac{z+1}{6}.$$

To handle this conveniently, we introduce a parameter, say $$t$$, and write

$$\frac{x-1}{2}=t,\qquad \frac{y-2}{3}=t,\qquad \frac{z+1}{6}=t.$$

From these three equalities we obtain the coordinates of an arbitrary point on the line:

$$x = 1 + 2t,$$

$$y = 2 + 3t,$$

$$z = -1 + 6t.$$

The point of intersection of this line with the plane must also satisfy the plane equation

$$2x - y + z = 6.$$

Substituting the expressions for $$x,\,y,\,z$$ obtained from the line into the plane, we get

$$2(1 + 2t) \;-\; (2 + 3t) \;+\; (-1 + 6t) = 6.$$

Expanding every term, we have

$$2 + 4t \;-\; 2 - 3t \;-\; 1 + 6t = 6.$$

Now we collect like terms. First, combine the constant terms:

$$2 - 2 - 1 = -1.$$

Next, combine the terms containing $$t$$:

$$4t - 3t + 6t = 7t.$$

So the left-hand side simplifies to

$$7t - 1.$$

Setting this equal to the right-hand side $$6$$ gives

$$7t - 1 = 6.$$

Adding $$1$$ to both sides, we obtain

$$7t = 7,$$

and dividing by $$7$$ yields

$$t = 1.$$

We now substitute $$t = 1$$ back into $$x = 1 + 2t,\, y = 2 + 3t,\, z = -1 + 6t$$ to locate the precise point of intersection, say $$P$$:

$$x_P = 1 + 2(1) = 1 + 2 = 3,$$

$$y_P = 2 + 3(1) = 2 + 3 = 5,$$

$$z_P = -1 + 6(1) = -1 + 6 = 5.$$

Hence the coordinates of the intersection point are

$$P(3,\,5,\,5).$$

We are asked for the square of the distance between this point $$P(3,5,5)$$ and the fixed point $$Q(-1,-1,2).$$

Recall the distance formula in three dimensions: for points $$(x_1,y_1,z_1)$$ and $$(x_2,y_2,z_2),$$

$$\text{distance}^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2.$$

Using $$P(3,5,5)$$ as $$(x_2,y_2,z_2)$$ and $$Q(-1,-1,2)$$ as $$(x_1,y_1,z_1),$$ we compute each difference:

$$x\text{-difference} = 3 - (-1) = 3 + 1 = 4,$$

$$y\text{-difference} = 5 - (-1) = 5 + 1 = 6,$$

$$z\text{-difference} = 5 - 2 = 3.$$

Squaring these differences gives

$$4^2 = 16,\qquad 6^2 = 36,\qquad 3^2 = 9.$$

Adding the squared differences, we find

$$16 + 36 + 9 = 61.$$

Thus the square of the distance between the required points is

$$61.$$

So, the answer is $$61$$.