The sum of the coordinates of one of the poin | vedclass

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(B) Let the general point on the line be $P(k + 2, -2k - 3, 2k - 5)$.The distance between $P$ and $(2, -3, -5)$ is given as $3$.Using the distance for

Vedclass · Accepted Answer

-$7$

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(B) Let the general point on the line be $P(k + 2, -2k - 3, 2k - 5)$.The distance between $P$ and $(2, -3, -5)$ is given as $3$.Using the distance formula: $\sqrt{(k + 2 - 2)^2 + (-2k - 3 + 3)^2 + (2k - 5 + 5)^2} = 3$.$\sqrt{k^2 + (-2k)^2 + (2k)^2} = 3$.$\sqrt{k^2 + 4k^2 + 4k^2} = 3$.$\sqrt{9k^2} = 3 \implies 3|k| = 3 \implies k = \pm 1$.If $k = 1$, the point is $P(1 + 2, -2(1) - 3, 2(1) - 5) = (3, -5, -3)$. Sum of coordinates $= 3 - 5 - 3 = -5$.If $k = -1$, the point is $P(-1 + 2, -2(-1) - 3, 2(-1) - 5) = (1, -1, -7)$. Sum of coordinates $= 1 - 1 - 7 = -7$.Since $-7$ is an option, the correct answer is $-7$.

The sum of the coordinates of one of the points on the line $\frac{x - 2}{1} = \frac{y + 3}{-2} = \frac{z + 5}{2}$ which is at a distance of $3 \text{ units}$ from the point $(2, -3, -5)$ is

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