The equation of the plane containing the line of intersection of $$2x - 5y + z = 3$$; $$x + y + 4z = 5$$, and parallel to the plane, $$x + 3y + 6z = 1$$, is

The equation of any plane passing through the line of intersection of the planes $$2x - 5y + z = 3$$ and $$x + y + 4z = 5$$ can be written as:

$$(2x - 5y + z - 3) + \lambda (x + y + 4z - 5) = 0$$

where $$\lambda$$ is a constant. Expanding this equation:

$$2x - 5y + z - 3 + \lambda x + \lambda y + 4\lambda z - 5\lambda = 0$$

Grouping the terms:

$$(2 + \lambda)x + (-5 + \lambda)y + (1 + 4\lambda)z + (-3 - 5\lambda) = 0$$

This plane must be parallel to the plane $$x + 3y + 6z = 1$$. For two planes to be parallel, their normal vectors must be proportional. The normal vector of the new plane is $$\langle 2 + \lambda, -5 + \lambda, 1 + 4\lambda \rangle$$, and the normal vector of the given plane is $$\langle 1, 3, 6 \rangle$$. Setting the ratios equal:

$$\frac{2 + \lambda}{1} = \frac{-5 + \lambda}{3} = \frac{1 + 4\lambda}{6}$$

Using the first two components:

$$\frac{2 + \lambda}{1} = \frac{-5 + \lambda}{3}$$

Cross-multiplying:

$$3(2 + \lambda) = 1 \cdot (-5 + \lambda)$$

$$6 + 3\lambda = -5 + \lambda$$

Bringing like terms together:

$$3\lambda - \lambda = -5 - 6$$

$$2\lambda = -11$$

$$\lambda = -\frac{11}{2}$$

Now, substituting $$\lambda = -\frac{11}{2}$$ back into the plane equation:

$$(2x - 5y + z - 3) + \left(-\frac{11}{2}\right)(x + y + 4z - 5) = 0$$

To eliminate the fraction, multiply the entire equation by 2:

$$2(2x - 5y + z - 3) - 11(x + y + 4z - 5) = 0$$

Expanding:

$$4x - 10y + 2z - 6 - 11x - 11y - 44z + 55 = 0$$

Combining like terms:

$$(4x - 11x) + (-10y - 11y) + (2z - 44z) + (-6 + 55) = 0$$

$$-7x - 21y - 42z + 49 = 0$$

Dividing the entire equation by $$-7$$ to simplify:

$$\frac{-7x}{-7} + \frac{-21y}{-7} + \frac{-42z}{-7} + \frac{49}{-7} = 0$$

$$x + 3y + 6z - 7 = 0$$

Thus, the equation is:

$$x + 3y + 6z = 7$$

Comparing with the options:

A. $$2x + 6y + 12z = -13$$

B. $$2x + 6y + 12z = 13$$

C. $$x + 3y + 6z = -7$$

D. $$x + 3y + 6z = 7$$

The equation matches option D. The plane $$x + 3y + 6z = 7$$ has the same normal vector $$\langle 1, 3, 6 \rangle$$ as the plane $$x + 3y + 6z = 1$$, so they are parallel. Since it was derived from the line of intersection, it contains that line.

Hence, the correct answer is Option D.