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The problem asks to find the distance from the point \( M_1(1; -1; -2) \) to a straight line given by the parametric form of the line equation:

\[
\frac{x + 3}{3} = \frac{y + 2}{2} = \frac{z - 8}{-2}
\]

Let's break this down into steps to find the solution.

### Step 1: Equation of the line
The given form of the line can be written as a parametric equation:

\[
x = -3 + 3t, \quad y = -2 + 2t, \quad z = 8 - 2t
\]

where \( t \) is a parameter. This gives the direction vector of the line as \( \mathbf{v} = (3, 2, -2) \) and the point on the line \( P_0(-3, -2, 8) \).

### Step 2: Distance formula from a point to a line in 3D
The distance \( d \) from a point \( M_1(x_1, y_1, z_1) \) to a line passing through \( P_0(x_0, y_0, z_0) \) and in the direction of \( \mathbf{v} = (a, b, c) \) is given by:

\[
d = \frac{| \mathbf{v} \times \overrightarrow{P_0 M_1} |}{|\mathbf{v}|}
\]

Where \( \overrightarrow{P_0 M_1} \) is the vector from \( P_0 \) to \( M_1 \), and \( \mathbf{v} \times \overrightarrow{P_0 M_1} \) is the cross product.

### Step 3: Apply the formula
The point \( M_1(1, -1, -2) \) and the point on the line \( P_0(-3, -2, 8) \) give us the vector \( \overrightarrow{P_0 M_1} = (1 - (-3), -1 - (-2), -2 - 8) = (4, 1, -10) \).

Now, compute the cross product of \( \mathbf{v} = (3, 2, -2) \) and \( \overrightarrow{P_0 M_1} = (4, 1, -10) \).

Next, calculate the magnitude of the cross product and the direction vector to get the distance.

I'll calculate this now.

M1(1; -1; -2) nuqtadan to'g'ri chiziqqacha bo'lgan masofani toping. Given line equation: (x+3)/3 = (y+2)/2 = (z-8)/-2.

Learn how to find the distance from point M1(1, -1, -2) to the line (x+3)/3 = (y+2)/2 = (z-8)/-2 in 3D space using the cross product and vector distance formula. Detailed solution provided, suitable for students in Grades 10-12.

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