Freedom equation of a line is (where are direction cosines)
`x = x_(1) + lr, y =y_(1) +mr, z=z_(1) + nr,` where:

Options

A. r is the distance of (x,y,z) from `(x_(1), y_(1), z_(1))`
B. `|r|` is the distance of (x,y,z) from `(x_(1),y_(1) ,z_(1))`
C. r is the distance of `(x_(1), y_(1), z_(1))` from the origin
D. none of these

Answer

To solve the problem regarding the freedom equation of a line given by the equations \( x = x_1 + lr \), \( y = y_1 + mr \), and \( z = z_1 + nr \), where \( l, m, n \) are the direction cosines, we will find the distance from the point \( (x_1, y_1, z_1) \) to a point \( (x, y, z) \) on the line.

### Step-by-Step Solution:

1. **Understand the Line Equation**:
The equations provided represent a line in three-dimensional space, where \( (x_1, y_1, z_1) \) is a point on the line and \( (l, m, n) \) are the direction cosines of the line.

2. **Distance Formula**:
The distance \( d \) between two points \( (x_1, y_1, z_1) \) and \( (x, y, z) \) is given by the formula:
\[ d = \sqrt{(x - x_1)^2 + (y - y_1)^2 + (z - z_1)^2} \]

3. **Substituting the Line Equations**:
Substitute \( x, y, z \) from the line equations into the distance formula:
\[ d = \sqrt{((x_1 + lr) - x_1)^2 + ((y_1 + mr) - y_1)^2 + ((z_1 + nr) - z_1)^2} \]
Simplifying this gives:
\[ d = \sqrt{(lr)^2 + (mr)^2 + (nr)^2} \]

4. **Factoring Out \( r^2 \)**:
We can factor out \( r^2 \) from the expression:
\[ d = \sqrt{r^2(l^2 + m^2 + n^2)} \]
This simplifies to:
\[ d = r \sqrt{l^2 + m^2 + n^2} \]

5. **Using the Property of Direction Cosines**:
Since \( l, m, n \) are direction cosines, we know that:
\[ l^2 + m^2 + n^2 = 1 \]
Therefore, substituting this into our distance formula gives:
\[ d = r \sqrt{1} = r \]

6. **Conclusion**:
Since \( r \) represents a distance, it must be non-negative. Thus, the distance \( d \) is equal to \( |r| \), but since \( r \) is a distance, we can conclude:
\[ d = r \]

### Final Answer:
The distance from the point \( (x_1, y_1, z_1) \) to the point \( (x, y, z) \) on the line is \( r \).