Point A lies at a distance of 6 units from th | vedclass

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(D) The given line is $\frac{x - 1}{2} = \frac{y}{2} = \frac{z - 1}{1} = k$. Any point on this line is given by $(2k + 1, 2k, k + 1)$. The distance be

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$(-3, -4, -1)$

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(D) The given line is $\frac{x - 1}{2} = \frac{y}{2} = \frac{z - 1}{1} = k$. Any point on this line is given by $(2k + 1, 2k, k + 1)$. The distance between this point and $(1, 0, 1)$ is $6$ units. Using the distance formula: $\sqrt{(2k + 1 - 1)^2 + (2k - 0)^2 + (k + 1 - 1)^2} = 6$. $\sqrt{(2k)^2 + (2k)^2 + k^2} = 6$. $\sqrt{4k^2 + 4k^2 + k^2} = 6$. $\sqrt{9k^2} = 6$. $3|k| = 6$,so $k = \pm 2$. Since the point lies in the negative $z$-direction relative to the point $(1, 0, 1)$,we look at the $z$-coordinate $z = k + 1$. For $k = -2$,$z = -2 + 1 = -1$,which is less than $1$. Substituting $k = -2$ into the coordinates: $x = 2(-2) + 1 = -3$,$y = 2(-2) = -4$,$z = -2 + 1 = -1$. Thus,the coordinates of $A$ are $(-3, -4, -1)$.

Point $A$ lies at a distance of $6$ units from the point $(1, 0, 1)$ on the line $\frac{x - 1}{2} = \frac{y}{2} = \frac{z - 1}{1}$ in the negative $z$-direction. Find the coordinates of $A$.

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