The square of the distance of the point ( $- 2, - 8,6$ ) from the line $\frac{x - 1}{1} = \frac{y -

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The square of the distance of the point ( $- 2, - 8,6$ ) from the line $\frac{x - 1}{1} = \frac{y - 1}{2} = \frac{z}{- 1}$ along the line $\frac{x + 5}{1} = \frac{y + 5}{- 1} = \frac{z}{2}$ is equal to:

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Sol. [Key Concept: Parametrise along the given direction, intersect the target line, then compute the distance.]DR or AB is $1, - 1,2$∴ Line AB is $\frac{x + 2}{1} = \frac{y + 8}{- 1} = \frac{z - 6}{2} = \lambda$B is ( $\lambda - 2, - \lambda - 8,2\lambda + 6$ )B is $(\mu + 1,2\mu + 1, - \mu)$Solving point B is $( - 3, - 7,4)$$${AB = \sqrt{( - 1)^{2} + (1)^{2} + (2)^{2}} = \sqrt{6} }{{AB}^{2} = 6 }$$

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