Find the coordinates of the foot of the perpendicular drawn from the origin \((0, 0, 0)\) to the straight line given by the equations \( \frac{x-2}{1} = \frac{y+1}{1} = \frac{z-3}{-1} \).
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To confirm your final answer, calculate the dot product of the calculated position vector \(\left(\frac{8}{3}, -\frac{1}{3}, \frac{7}{3}\right)\) with the line's direction vector \((1, 1, -1)\):
\[
\frac{8}{3}(1) + \left(-\frac{1}{3}\right)(1) + \frac{7}{3}(-1) = \frac{8 - 1 - 7}{3} = 0
\]
Since it evaluates exactly to 0, your calculations are perfectly correct!