Equation of the line, which is perpendicular to the line, $3 x+y=\lambda(\lambda \neq 0)$ and passing through origin, is given by
$$
\frac{x-0}{3}=\frac{y-0}{1}=r
$$
For foot of perpendicular
$$
r=\frac{-((3 \times 0)+(1 \times 0)-\lambda)}{3^2+1^2}=\frac{\lambda}{10}
$$
So, foot of perpendicular $P=\left(\frac{3 \lambda}{10}, \frac{\lambda}{10}\right)$
Given the line meets $\mathrm{X}$-axis at $A=\left(\frac{\lambda}{3}, 0\right)$ and meets $Y$-axis at $B=(0, \lambda)$
So,
$$
\begin{aligned}
&B P=\sqrt{\left(\frac{3 \lambda}{10}\right)^2+\left(\frac{\lambda}{10}-\lambda\right)^2} \Rightarrow B P=\sqrt{\frac{9 \lambda^2}{100}+\frac{81 \lambda^2}{100}} \\
&\Rightarrow B P=\sqrt{\frac{90 \lambda^2}{100}} \\
&\text { Now, } P A=\sqrt{\left(\frac{\lambda}{3}-\frac{3 \lambda}{10}\right)^2+\left(0-\frac{\lambda}{10}\right)^2} \\
&\Rightarrow P A \sqrt{\frac{\lambda^2}{900}+\frac{\lambda^2}{100}} \Rightarrow P A=\sqrt{\frac{10 \lambda^2}{900}}
\end{aligned}
$$
Therefore $B P: P A=3: 1$