Let the lines L_1: vec{r}=hat{i}+2hat{j}+3hat | vedclass

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(B) To find the point of intersection $R$, we equate the expressions for $\vec{r}$:$2\lambda+1 = 5\mu+4$, $3\lambda+2 = 2\mu+1$, $4\lambda+3 = \mu$.So

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$360$

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(B) To find the point of intersection $R$, we equate the expressions for $\vec{r}$:$2\lambda+1 = 5\mu+4$, $3\lambda+2 = 2\mu+1$, $4\lambda+3 = \mu$.Solving these, we get $\lambda = -1$ and $\mu = -1$.Thus, the point $R$ is $(-1, -1, -1)$.Point $P$ lies on $L_1$, so $P = (2\lambda+1, 3\lambda+2, 4\lambda+3)$.Given $|\overrightarrow{PR}| = \sqrt{29}$, so $PR^2 = 29$.$(2\lambda+1 - (-1))^2 + (3\lambda+2 - (-1))^2 + (4\lambda+3 - (-1))^2 = 29$.$(2\lambda+2)^2 + (3\lambda+3)^2 + (4\lambda+4)^2 = 29$.$29(\lambda+1)^2 = 29 \Rightarrow (\lambda+1)^2 = 1 \Rightarrow \lambda = 0$ or $\lambda = -2$.If $\lambda = 0$, $P = (1, 2, 3)$ (First octant). If $\lambda = -2$, $P = (-3, -4, -5)$ (Not in first octant).So, $P = (1, 2, 3)$.Point $Q$ lies on $L_2$, so $Q = (5\mu+4, 2\mu+1, \mu)$.Given $|\overrightarrow{PQ}|^2 = \frac{47}{3}$.$(5\mu+4-1)^2 + (2\mu+1-2)^2 + (\mu-3)^2 = \frac{47}{3}$.$(5\mu+3)^2 + (2\mu-1)^2 + (\mu-3)^2 = \frac{47}{3}$.$25\mu^2 + 30\mu + 9 + 4\mu^2 - 4\mu + 1 + \mu^2 - 6\mu + 9 = \frac{47}{3}$.$30\mu^2 + 20\mu + 19 = \frac{47}{3} \Rightarrow 90\mu^2 + 60\mu + 57 = 47 \Rightarrow 90\mu^2 + 60\mu + 10 = 0$.$9\mu^2 + 6\mu + 1 = 0 \Rightarrow (3\mu+1)^2 = 0 \Rightarrow \mu = -\frac{1}{3}$.$Q = (5(-\frac{1}{3})+4, 2(-\frac{1}{3})+1, -\frac{1}{3}) = (\frac{7}{3}, \frac{1}{3}, -\frac{1}{3})$.$(QR)^2 = (\frac{7}{3} - (-1))^2 + (\frac{1}{3} - (-1))^2 + (-\frac{1}{3} - (-1))^2 = (\frac{10}{3})^2 + (\frac{4}{3})^2 + (\frac{2}{3})^2 = \frac{100+16+4}{9} = \frac{120}{9}$.$27(QR)^2 = 27 	imes \frac{120}{9} = 3 	imes 120 = 360$.

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