If the points $$(1, 1, \lambda)$$ and $$(-3, 0, 1)$$, are equidistant from the plane, $$3x + 4y - 12z + 13 = 0$$, then $$\lambda$$ satisfies the equation:

The problem states that the points $$(1, 1, \lambda)$$ and $$(-3, 0, 1)$$ are equidistant from the plane $$3x + 4y - 12z + 13 = 0$$. We need to find the equation that $$\lambda$$ satisfies.

The distance from a point $$(x_1, y_1, z_1)$$ to a plane $$ax + by + cz + d = 0$$ is given by:

$$\text{Distance} = \frac{|ax_1 + by_1 + cz_1 + d|}{\sqrt{a^2 + b^2 + c^2}}$$

For the plane $$3x + 4y - 12z + 13 = 0$$, we have $$a = 3$$, $$b = 4$$, $$c = -12$$, and $$d = 13$$. The denominator is:

$$\sqrt{3^2 + 4^2 + (-12)^2} = \sqrt{9 + 16 + 144} = \sqrt{169} = 13$$

Now, the distance from point $$A(1, 1, \lambda)$$ to the plane is:

$$d_1 = \frac{|3(1) + 4(1) + (-12)(\lambda) + 13|}{13} = \frac{|3 + 4 - 12\lambda + 13|}{13} = \frac{|20 - 12\lambda|}{13}$$

The distance from point $$B(-3, 0, 1)$$ to the plane is:

$$d_2 = \frac{|3(-3) + 4(0) + (-12)(1) + 13|}{13} = \frac{|-9 - 12 + 13|}{13} = \frac{|-8|}{13} = \frac{8}{13}$$

Since the points are equidistant, $$d_1 = d_2$$:

$$\frac{|20 - 12\lambda|}{13} = \frac{8}{13}$$

Multiplying both sides by 13 to clear the denominator:

$$|20 - 12\lambda| = 8$$

An absolute value equation $$|A| = B$$ (with $$B > 0$$) means $$A = B$$ or $$A = -B$$. So, we have two cases:

Case 1: $$20 - 12\lambda = 8$$

Subtract 20 from both sides:

$$-12\lambda = 8 - 20$$

$$-12\lambda = -12$$

Divide both sides by -12:

$$\lambda = \frac{-12}{-12} = 1$$

Case 2: $$20 - 12\lambda = -8$$

Subtract 20 from both sides:

$$-12\lambda = -8 - 20$$

$$-12\lambda = -28$$

Divide both sides by -12:

$$\lambda = \frac{-28}{-12} = \frac{28}{12} = \frac{7}{3}$$

So, $$\lambda$$ can be either $$1$$ or $$\frac{7}{3}$$. We need a quadratic equation that has these roots. The quadratic equation with roots $$\alpha$$ and $$\beta$$ is:

$$x^2 - (\alpha + \beta)x + \alpha\beta = 0$$

Here, $$\alpha = 1$$ and $$\beta = \frac{7}{3}$$. The sum of roots is:

$$\alpha + \beta = 1 + \frac{7}{3} = \frac{3}{3} + \frac{7}{3} = \frac{10}{3}$$

The product of roots is:

$$\alpha\beta = 1 \times \frac{7}{3} = \frac{7}{3}$$

Substituting into the quadratic form:

$$x^2 - \frac{10}{3}x + \frac{7}{3} = 0$$

To eliminate fractions, multiply every term by 3:

$$3x^2 - 10x + 7 = 0$$

Since the variable is $$\lambda$$, we write:

$$3\lambda^2 - 10\lambda + 7 = 0$$

Comparing with the options:

A. $$3\lambda^2 + 10\lambda + 7 = 0$$

B. $$3\lambda^2 + 10x - 13 = 0$$ (contains $$x$$, likely a typo)

C. $$3\lambda^2 - 10\lambda + 7 = 0$$

D. $$3\lambda^2 - 10\lambda + 21 = 0$$

Option C matches the equation we derived.

Hence, the correct answer is Option C.