Let $$(\lambda, 2, 1)$$ be a point on the plane which passes through the point $$(4, -2, 2)$$. If the plane is perpendicular to the line joining the points $$(-2, -21, 29)$$ and $$(-1, -16, 23)$$, then $$\left(\frac{\lambda}{11}\right)^2 - \frac{4\lambda}{11} - 4$$ is equal to ______.

The plane passes through $$(4, -2, 2)$$ and is perpendicular to the line joining $$(-2, -21, 29)$$ and $$(-1, -16, 23)$$.

The direction vector of the line is $$(-1-(-2), -16-(-21), 23-29) = (1, 5, -6)$$. Since the plane is perpendicular to this line, $$(1, 5, -6)$$ is the normal vector of the plane.

The equation of the plane passing through $$(4, -2, 2)$$ with normal $$(1, 5, -6)$$ is:

$$1(x-4) + 5(y+2) - 6(z-2) = 0$$

$$x - 4 + 5y + 10 - 6z + 12 = 0$$

$$x + 5y - 6z + 18 = 0$$

The point $$(\lambda, 2, 1)$$ lies on this plane, so:

$$\lambda + 5(2) - 6(1) + 18 = 0$$

$$\lambda + 10 - 6 + 18 = 0$$

$$\lambda + 22 = 0$$, so $$\lambda = -22$$.

Now we compute $$\left(\frac{\lambda}{11}\right)^2 - \frac{4\lambda}{11} - 4 = \left(\frac{-22}{11}\right)^2 - \frac{4(-22)}{11} - 4 = (-2)^2 + 8 - 4 = 4 + 8 - 4 = 8$$.