Answer
$\lambda_3\lt \lambda_2\lt \lambda_1$
Work Step by Step
According to Doppler's effect
$f_{\circ}=f_s[\frac{1+(\frac{v_{\circ}}{v})}{1-(\frac{v_s}{v})}]$...eq(1)
We plug in the known values to obtain:
$f_2=f_1[\frac{1+\frac{0}{v}}{1-\frac{u}{v}}]$
$\implies f_2=\frac{f_1}{1-\frac{u}{v}}$
As $u\lt v$
$\implies 0\lt 1-\frac{u}{v}\lt 1$
Thus, $f_2\gt f_1$
Now plug in $u$ for $v_{\circ}$, $0$ for $v_s$ and $f_2$ for $f_s$ in eq(1) to obtain:
$f_3=f_2[\frac{1+(\frac{u}{v})}{1-(\frac{0}{v})}]$
$f_3=(1+\frac{u}{v})f_2$
This shows that $f_3\gt f_2$
Thus, $f_3\gt f_2\gt f_1$
As frequency is inversely proportional to the wavelength, therefore the order of the wavelength is $\lambda_3\lt \lambda_2\lt \lambda_1$
