Let L be the line frac{x+1}{2}=frac{y+1}{3}=f | vedclass

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(A) Let any point on the line $L$ be $(a, b, c) = (2k-1, 3k-1, 6k-3)$.Given that the distance of this point from the point $P(-1, -1, -9)$ is $7$.Usin

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$34$

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(A) Let any point on the line $L$ be $(a, b, c) = (2k-1, 3k-1, 6k-3)$.Given that the distance of this point from the point $P(-1, -1, -9)$ is $7$.Using the distance formula: $\sqrt{(2k-1 - (-1))^2 + (3k-1 - (-1))^2 + (6k-3 - (-9))^2} = 7$.$\sqrt{(2k)^2 + (3k)^2 + (6k+6)^2} = 7$.Squaring both sides: $4k^2 + 9k^2 + (6k+6)^2 = 49$.$13k^2 + 36k^2 + 72k + 36 = 49$.$49k^2 + 72k - 13 = 0$.This is a quadratic equation in $k$. Let the roots be $k_1$ and $k_2$.The sum of the coordinates for a point $(a, b, c)$ is $a+b+c = (2k-1) + (3k-1) + (6k-3) = 11k - 5$.For the two points in $S$, the sum is $(11k_1 - 5) + (11k_2 - 5) = 11(k_1 + k_2) - 10$.From the quadratic equation, $k_1 + k_2 = -\frac{72}{49}$.Sum $= 11(-\frac{72}{49}) - 10 = -\frac{792}{49} - 10 = -\frac{1282}{49}$.

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 point $P(-1, -1, -9)$ is $7$. Then $\sum_{(a,b,c)\in S} (a+b+c)$ is equal to:

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