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Let the given line be $L$. A general point on the line $L$ can be represented as: $$ \frac{x+5}{1}=\frac{y+3}{4}=\frac{z-6}{-9} = \lambda $$ So, $x = \lambda - 5$, $y = 4\lambda - 3$, and $z = -9\lambda + 6$. Thus, any point $Q$ on the line $L$ can be written as $Q(\lambda - 5, 4\lambda - 3, -9\lambda + 6)$.
Let $P(2,4,-1)$ be the given point. The direction ratios of the line segment $PQ$ are given by: $$ (\lambda - 5 - 2, 4\lambda - 3 - 4, -9\lambda + 6 - (-1)) = (\lambda - 7, 4\lambda - 7, -9\lambda + 7) $$
Since we want to find the shortest distance, $PQ$ must be perpendicular to the line $L$. The direction ratios of line $L$ are $(1, 4, -9)$. For $PQ$ to be perpendicular to $L$, the dot product of their direction ratios must be zero: $$ 1(\lambda - 7) + 4(4\lambda - 7) - 9(-9\lambda + 7) = 0 $$ $$ \lambda - 7 + 16\lambda - 28 + 81\lambda - 63 = 0 $$ $$ 98\lambda - 98 = 0 $$ $$ \lambda = 1 $$
Substitute $\lambda = 1$ into the coordinates of point $Q$: $$ Q(1 - 5, 4(1) - 3, -9(1) + 6) = Q(-4, 1, -3) $$
Now, find the distance between $P(2,4,-1)$ and $Q(-4,1,-3)$: $$ PQ = \sqrt{(-4 - 2)^2 + (1 - 4)^2 + (-3 - (-1))^2} $$ $$ PQ = \sqrt{(-6)^2 + (-3)^2 + (-2)^2} $$ $$ PQ = \sqrt{36 + 9 + 4} $$ $$ PQ = \sqrt{49} $$ $$ PQ = 7 $$
Final Answer: 7