Step 1: The original expression We are given the statement \( \sim [p \vee (\sim (p \land q))] \). We need to simplify this expression and find the equivalent logical statement.
Step 2: Apply De Morgan's law First, apply De Morgan’s law to the negation of the disjunction \( \sim [p \vee (\sim (p \land q))] \). De Morgan’s law states that \( \sim (A \vee B) = \sim A \land \sim B \), so we get: \[ \sim p \land \sim (\sim (p \land q)) \]
Step 3: Simplify the inner negation Now, simplify the double negation \( \sim (\sim (p \land q)) \), which cancels out the two negations, giving us: \[ \sim p \land (p \land q) \]
Step 4: Conclusion Thus, the expression simplifies to: \[ (p \land q) \land (\sim p) \] which is the correct equivalent form of the original expression.
Consider the parabola \(25[(x-2)^2 + (y+5)^2] = (3x+4y-1)^2\), match the characteristic of this parabola given in List-I with its corresponding item in List-II.

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}