Step 1: Understanding the decay process. Radioactive decay follows the equation: \[ N = N_0 \times \left( \frac{1}{2} \right)^n \] Where \( N_0 \) is the initial number of nuclei, \( N \) is the number of nuclei left after \( n \) half-lives, and \( n \) is the number of half-lives elapsed. If \( \frac{7}{8} \) of the substance has disintegrated, then \( \frac{1}{8} \) of the substance remains. This means the number of remaining radioactive nuclei is \( \frac{N_0}{8} \).
Step 2: Finding the number of half-lives. Using the decay equation: \[ \frac{N_0}{2^n} = \frac{N_0}{8} \] Simplifying: \[ 2^n = 8 \] \[ n = 3 \] So, 3 half-lives have elapsed.
Step 3: Calculating the time. Since the half-life is \( T \), the time taken for the disintegration of \( \frac{7}{8} \) of the substance is: \[ \text{Time} = 3 \times T = 3T \]


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}