Question
Let L be the line $\frac{x + 1}{2} = \frac{y + 1}{3} = \frac{z + 3}{6}$ and let S be the set of all points $(a,b,c)$ on L , whose distance from the line $\frac{x + 1}{2} = \frac{y + 1}{3} = \frac{z - 9}{0}$ along the line L is 7 . Then $\sum_{(a,b,c) \in S}\mspace{2mu}(a + b + c)$ is equal to :
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Solution
$M$ is the point of intersection of $L_{1}\& L_{2}$
$${\Rightarrow 2\lambda - 1 = 2\mu - 1,3\lambda - 1 = 3\mu - 1,6\lambda - 3 = 9 }{\Rightarrow \lambda = 2 = \mu }{\Rightarrow M(3,5,9) }$$Now let point P be $(2\text{ }K - 1,3\text{ }K - 1,6\text{ }K - 3)$ on $L_{2}$ such that $PM = 7$
$${\Rightarrow \sqrt{(2\text{ }K - 4)^{2} + (3\text{ }K - 6)^{2} + (6\text{ }K - 12)^{2}} = 7 }{\Rightarrow 49{\text{ }K}^{2} + 196 - 196\text{ }K = 49 }{\Rightarrow K^{2} + 4 - 4\text{ }K = 1 }{\Rightarrow K^{2} - 4\text{ }K + 3 = 0 }{\Rightarrow K = 1,3 }$$So points $P\& Q$ are $(1,2,3)\&(5,8,15)$
So sum of all co-ordinates of P & Q = 34
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