Distance of the point $(2, 4, -1)$ from the line $\frac{10+2x}{2} = \frac{y+3}{4} = \frac{6-z}{9}$ is

The line is given as $\frac{10+2x}{2} = \frac{y+3}{4} = \frac{6-z}{9}$

Simplifying:

$\frac{10+2x}{2} = \frac{2(5+x)}{2} = 5+x = \frac{x+5}{1}$

$\frac{6-z}{9} = \frac{-(z-6)}{9} = \frac{z-6}{-9}$

The line can be written as:

$\frac{x+5}{1} = \frac{y+3}{4} = \frac{z-6}{-9}$

A point on the line is $A = (-5, -3, 6)$ and the direction vector is $\vec{b} = (1, 4, -9)$

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The vector from point $A$ to point $P(2, 4, -1)$ is:

$\vec{AP} = P - A$

$\vec{AP} = (2, 4, -1) - (-5, -3, 6)$

$\vec{AP} = (7, 7, -7)$

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The cross product $\vec{AP} \times \vec{b}$:

$\vec{AP} \times \vec{b} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 7 & 7 & -7 \\ 1 & 4 & -9 \end{vmatrix}$

$\vec{i}$ component: $7(-9) - (-7)(4) = -63 + 28 = -35$

$\vec{j}$ component: $-[7(-9) - (-7)(1)] = -(-63 + 7) = 56$

$\vec{k}$ component: $7(4) - 7(1) = 28 - 7 = 21$

Therefore $\vec{AP} \times \vec{b} = (-35, 56, 21)$

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The magnitude of the cross product:

$|\vec{AP} \times \vec{b}| = \sqrt{(-35)^2 + 56^2 + 21^2}$

$= \sqrt{1225 + 3136 + 441}$

$= \sqrt{4802}$

$= \sqrt{2 \times 49^2}$

$= 49\sqrt{2}$

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The magnitude of the direction vector:

$|\vec{b}| = \sqrt{1^2 + 4^2 + (-9)^2}$

$= \sqrt{1 + 16 + 81}$

$= \sqrt{98}$

$= 7\sqrt{2}$

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The distance from the point to the line:

$d = \frac{|\vec{AP} \times \vec{b}|}{|\vec{b}|}$

$= \frac{49\sqrt{2}}{7\sqrt{2}}$

$= 7$

The distance is $7$ units.