Answer
See the detailed answer below.
Work Step by Step
We know, in diffraction gratings, that the angle $\theta$ increases when the wavlelngth increases.
$$d\sin\theta=m\lambda$$
And to see, or to proof, that the second and the third order of spectra of white light produced by a diffraction grating is overlapping, we need to find the angle of the second and the third spectra.
Thus,
$$\sin\theta=\dfrac{m\lambda}{d}$$
Now we need to find the ratio of the second order and third order.
$$\dfrac{\sin\theta_2}{\sin\theta_3}=\dfrac{\dfrac{2\lambda_2}{d}}{\dfrac{3\lambda_3}{d}}=\dfrac{2\lambda_2}{3\lambda_3}$$
Plugging the known,
$$\dfrac{\sin\theta_2}{\sin\theta_3} =\dfrac{2\cdot 700 }{3\cdot 400}=1.17$$
This means that,
$$ \sin\theta_2 =1.17\sin\theta_3$$
This means that the maximum angle of the second order is greater than the minimum angle of the third order; so that we have an overlap.
To find which wavelengths overlap, we need to assume that the two angles are equal, so that $\dfrac{\sin\theta_2}{\sin\theta_3}=1$.
Thus,
$$1=\dfrac{2\lambda_2}{3\lambda_3}$$
And hence,
$$\lambda_3=\dfrac{2\lambda_{2,maximum}}{3}=\dfrac{2\cdot 700}{3}=\bf 466.67\;\rm nm$$
$$\lambda_2=\dfrac{3\lambda_{3,minimum}}{2}=\dfrac{3\cdot 400}{2}=\bf 600\;\rm nm$$
It is obvious now that the two wavelengths of 600 nm and 700 nm overlapping with the two wavelengths of 400 nm and 467 nm.