The foot of the perpendicular drawn from the origin, on the line, $$3x + y = \lambda(\lambda \neq 0)$$ is P. If the line meets x-axis at A and y-axis at B, then the ratio BP : PA is:

The given line is $$3x + y = \lambda$$, where $$\lambda \neq 0$$.

The line meets the x-axis ($$y = 0$$) at $$A = \left(\frac{\lambda}{3}, 0\right)$$ and the y-axis ($$x = 0$$) at $$B = (0, \lambda)$$.

The foot of the perpendicular from the origin $$O(0,0)$$ to the line $$3x + y = \lambda$$ is point $$P$$.

The line $$3x + y = \lambda$$ can be written as $$3x + y - \lambda = 0$$. Using the formula for the foot of the perpendicular from $$(x_0, y_0)$$ to $$ax + by + c = 0$$:

$$\frac{x - x_0}{a} = \frac{y - y_0}{b} = -\frac{ax_0 + by_0 + c}{a^2 + b^2}$$

Here $$a = 3$$, $$b = 1$$, $$c = -\lambda$$, and $$(x_0, y_0) = (0, 0)$$:

$$\frac{x}{3} = \frac{y}{1} = -\frac{0 + 0 - \lambda}{9 + 1} = \frac{\lambda}{10}$$

So $$P = \left(\frac{3\lambda}{10}, \frac{\lambda}{10}\right)$$.

Now we find the distances $$BP$$ and $$PA$$.

$$BP^2 = \left(\frac{3\lambda}{10} - 0\right)^2 + \left(\frac{\lambda}{10} - \lambda\right)^2 = \frac{9\lambda^2}{100} + \frac{81\lambda^2}{100} = \frac{90\lambda^2}{100}$$

$$BP = \frac{3\lambda\sqrt{10}}{10}$$

$$PA^2 = \left(\frac{\lambda}{3} - \frac{3\lambda}{10}\right)^2 + \left(0 - \frac{\lambda}{10}\right)^2 = \left(\frac{10\lambda - 9\lambda}{30}\right)^2 + \frac{\lambda^2}{100} = \frac{\lambda^2}{900} + \frac{\lambda^2}{100} = \frac{\lambda^2 + 9\lambda^2}{900} = \frac{10\lambda^2}{900}$$

$$PA = \frac{\lambda\sqrt{10}}{30}$$

Therefore: $$\frac{BP}{PA} = \frac{\frac{3\lambda\sqrt{10}}{10}}{\frac{\lambda\sqrt{10}}{30}} = \frac{3}{10} \times 30 = 9$$

So $$BP : PA = 9 : 1$$.

The answer is Option A: 9 : 1.