To determine the correctness of the given statements, let's analyze each one individually:
By reviewing both statements, it is clear that both Statement I and Statement II are true. Therefore, the correct answer is: Both Statement I and Statement II are true.
- Statement I: Transition metals tend to have more stable higher oxidation states as we move down the group due to the increasing availability of d-orbitals. This is in contrast to p-block elements, where higher oxidation states become less stable down the group.
- Statement II: Copper can indeed liberate hydrogen from weak acids such as hydrochloric acid, which is true for copper in the form of copper(I) chloride.
Thus, both statements are true, and the correct answer is (3).
The given circuit works as: 
Let the lines $L_1 : \vec r = \hat i + 2\hat j + 3\hat k + \lambda(2\hat i + 3\hat j + 4\hat k)$, $\lambda \in \mathbb{R}$ and $L_2 : \vec r = (4\hat i + \hat j) + \mu(5\hat i + + 2\hat j + \hat k)$, $\mu \in \mathbb{R}$ intersect at the point $R$. Let $P$ and $Q$ be the points lying on lines $L_1$ and $L_2$, respectively, such that $|PR|=\sqrt{29}$ and $|PQ|=\sqrt{\frac{47}{3}}$. If the point $P$ lies in the first octant, then $27(QR)^2$ is equal to}