Question
Let the lines $L_{1}:\overrightarrow{r} = \widehat{i} + 2\widehat{j} + 3\widehat{k} + \lambda(2\widehat{i} + 3\widehat{j} + 4\widehat{k})$, $\lambda \in \mathbb{R}$ and $L_{2}:\overrightarrow{r} = (4\widehat{i} + \widehat{j}) + \mu(5\widehat{i} + 2\widehat{j} + \widehat{k}),\mu \in \mathbb{R}$, intersect at the point R . Let P and Q be the points lying on lines $L_{1}$ and $L_{2}$, respectively, such that $|\overrightarrow{PR}| = \sqrt{29}$ and $|\overrightarrow{PQ}| = \sqrt{\frac{47}{3}}$. If the point P lies in the first octant, then $27(QR)^{2}$ is equal to
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Solution
For POI
$$\begin{matrix} & 2\lambda + 1 = 5\mu + 4;3\lambda + 2 = 2\mu + 1;4\lambda + 3 = \mu \\ & \ \Rightarrow \lambda = \mu = - 1 \\ & R( - 1, - 1, - 1)\ P(2\lambda + 1,3\lambda + 2,4\lambda + 3) \\ & {PR}^{2} = 29 \Rightarrow (2\lambda + 2)^{2} + (3\lambda + 3)^{2} + (4\lambda + 4)^{2} = 29 \\ & \ \Rightarrow \lambda = 0\text{~}\text{or}\text{~}\lambda = - 2\text{~}\text{(Reject)}\text{~} \\ & \ \Rightarrow P(1,2,3) \\ & Q(5\mu + 4,2\mu + 1,\mu) \\ & \ |PQ| = \sqrt{\frac{47}{3}} \Rightarrow {PQ}^{2} = \frac{47}{3} \\ & \ \Rightarrow (5\mu + 3)^{2} + (2\mu - 1)^{2} + (\mu - 3)^{2} = \frac{47}{3} \\ & \ \Rightarrow \mu = - \frac{1}{3} \\ & Q = \left( \frac{7}{3},\frac{1}{3}, - \frac{1}{3} \right) \\ & \ (QR)^{2} = \left( \frac{7}{3} + 1 \right)^{2} + \left( \frac{1}{3} + 1 \right)^{2} + \left( - \frac{1}{3} + 1 \right)^{2} \\ & \ = \frac{100 + 16 + 4}{9} = \frac{120}{9} \\ & \ \Rightarrow 27 \times (QR)^{2} = 27 \times \frac{120}{9} = 360 \end{matrix}$$
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