Given forces

The two force vectors are $\vec A = \vec P + \vec Q, \qquad \vec B = \vec P - \vec Q$ with $\vec P \perp \vec Q$

Magnitudes of the two forces

Since $\vec P \perp \vec Q$, $|\vec A|^2 = |\vec P + \vec Q|^2 = P^2 + Q^2$ $|\vec B|^2 = |\vec P - \vec Q|^2 = P^2 + Q^2$

Hence, $|\vec A| = |\vec B| = \sqrt{P^2+Q^2}$

Resultant of two equal-magnitude vectors making angle $\theta$

If two vectors of magnitude $R$ make angle $\theta$, then resultant magnitude is $\sqrt{R^2+R^2+2R^2\cos\theta}=R\sqrt{2(1+\cos\theta)}$

Here $R=\sqrt{P^2+Q^2}$, so resultant magnitude becomes $\mathcal R = \sqrt{2(P^2+Q^2)(1+\cos\theta)}$

For angle $\theta_1$

Given resultant magnitude is $\sqrt{3(P^2+Q^2)}$

So, $2(P^2+Q^2)(1+\cos\theta_1)=3(P^2+Q^2)$ $2(1+\cos\theta_1)=3$ $1+\cos\theta_1=\frac32$ $\cos\theta_1=\frac12$

Therefore, $\theta_1=60^\circ$

For angle $\theta_2$

Given resultant magnitude is $\sqrt{2(P^2+Q^2)}$

So, $2(P^2+Q^2)(1+\cos\theta_2)=2(P^2+Q^2)$ $1+\cos\theta_2=1$ $\cos\theta_2=0$

Therefore, $\theta_2=90^\circ$

Check Statement I

Statement I says this is possible only when $\theta_1<\theta_2$

Since $60^\circ<90^\circ$ Statement I is true.

Check Statement II

Statement II says $\theta_1=60^\circ, \qquad \theta_2=90^\circ$

This matches our result exactly, so Statement II is true.

Final choice

Both Statement I and Statement II are true.

Therefore, the correct option is B.