(P \(\lor\) R) \(\Rightarrow\) Q}
(P \(\Rightarrow\) R) \(\lor\) (Q \(\Rightarrow\) R)}
(P \(\Rightarrow\) R) \(\land\) (Q \(\Rightarrow\) R)
(P \(\land\) R) \(\Rightarrow\) Q
We are given the statement (P \(\Rightarrow\) Q) \(\land\) (R \(\Rightarrow\) Q). We know that P \(\Rightarrow\) Q is equivalent to \(\neg\)P \(\lor\) Q. So, the given statement can be written as: \[ (\neg P \lor Q) \land (\neg R \lor Q). \] Using the distributive law, we can rewrite this as: \[ (\neg P \land \neg R) \lor Q. \] Using De Morgan's law, \(\neg P \land \neg R\) is equivalent to \(\neg(P \lor R)\). So the statement becomes: \[ \neg(P \lor R) \lor Q. \] This is equivalent to (P \(\lor\) R) \(\Rightarrow\) Q.
The heat generated in 1 minute between points A and B in the given circuit, when a battery of 9 V with internal resistance of 1 \(\Omega\) is connected across these points is ______ J. 
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}