Let $$P$$ be a plane $$lx + my + nz = 0$$ containing the line, $$\frac{1-x}{1} = \frac{y+4}{2} = \frac{z+2}{3}$$. If plane $$P$$ divides the line segment $$AB$$ joining points $$A(-3, -6, 1)$$ and $$B(2, 4, -3)$$ in ratio $$k:1$$ then the value of $$k$$ is equal to:

The line is $$\frac{1-x}{1} = \frac{y+4}{2} = \frac{z+2}{3}$$, which can be rewritten as $$\frac{x-1}{-1} = \frac{y+4}{2} = \frac{z+2}{3}$$. This line passes through $$(1, -4, -2)$$ with direction ratios $$(-1, 2, 3)$$.

Since the plane $$lx + my + nz = 0$$ contains this line, the point $$(1, -4, -2)$$ lies on the plane: $$l - 4m - 2n = 0$$. Also, the direction $$(-1, 2, 3)$$ is perpendicular to the normal $$(l, m, n)$$: $$-l + 2m + 3n = 0$$.

Adding these two equations: $$-2m + n = 0$$, so $$n = 2m$$. From the first equation: $$l = 4m + 2n = 4m + 4m = 8m$$. Taking $$m = 1$$: $$l = 8, m = 1, n = 2$$.

The plane is $$8x + y + 2z = 0$$.

The line segment $$AB$$ joins $$A(-3, -6, 1)$$ and $$B(2, 4, -3)$$. The point dividing $$AB$$ in ratio $$k:1$$ is $$\left(\frac{2k - 3}{k+1}, \frac{4k - 6}{k+1}, \frac{-3k + 1}{k+1}\right)$$.

Substituting into the plane equation: $$8\left(\frac{2k-3}{k+1}\right) + \left(\frac{4k-6}{k+1}\right) + 2\left(\frac{-3k+1}{k+1}\right) = 0$$.

Simplifying the numerator: $$8(2k-3) + (4k-6) + 2(-3k+1) = 16k - 24 + 4k - 6 - 6k + 2 = 14k - 28 = 0$$.

Therefore $$k = 2$$, which corresponds to option (3).