For positive l, m and n, if the points `x=ny+mz, y=lz+nx, z=mx+ly` intersect in a straight line, when
Q. l, m and n satisfy the equation

Options

A. `l^2+m^2+n^2=2`
B. `l^2+m^2+n^2+2mln=1`
C. `l^2+m^2+n^2=1`
D. None of these

Answer

To solve the problem, we need to analyze the given equations and find the condition under which the points defined by these equations intersect in a straight line. The equations given are:

1. \( x = ny + mz \)
2. \( y = lz + nx \)
3. \( z = mx + ly \)

### Step 1: Rewrite the equations
We can rearrange each equation to bring all terms to one side:

1. \( x - ny - mz = 0 \) (Equation 1)
2. \( nx - y + lz = 0 \) (Equation 2)
3. \( mx + ly - z = 0 \) (Equation 3)

### Step 2: Identify the direction vectors
The direction vectors of the lines represented by these equations can be derived from the coefficients of \(x\), \(y\), and \(z\) in each equation.

- From Equation 1: The direction vector is \( \mathbf{d_1} = (1, -n, -m) \)
- From Equation 2: The direction vector is \( \mathbf{d_2} = (n, -1, l) \)
- From Equation 3: The direction vector is \( \mathbf{d_3} = (m, l, -1) \)

### Step 3: Find the cross product of direction vectors
To find the condition for the lines to intersect, we need to compute the cross product of two of these direction vectors and ensure that the third vector lies in the same plane.

Let's compute the cross product \( \mathbf{d_1} \times \mathbf{d_2} \):

\[ \mathbf{d_1} \times \mathbf{d_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & -n & -m \\ n & -1 & l \end{vmatrix} \]

Calculating the determinant, we get:

\[ \mathbf{d_1} \times \mathbf{d_2} = \mathbf{i}((-n)(l) - (-m)(-1)) - \mathbf{j}(1 \cdot l - (-m)(n)) + \mathbf{k}(1 \cdot (-1) - (-n)(n)) \]

This simplifies to:

\[ \mathbf{d_1} \times \mathbf{d_2} = (-nl + m) \mathbf{i} - (l + mn) \mathbf{j} + (n^2 - 1) \mathbf{k} \]

### Step 4: Check if the third direction vector lies in the plane
For the lines to intersect, the third direction vector \( \mathbf{d_3} = (m, l, -1) \) must be orthogonal to the normal vector obtained from the cross product. This means:

\[ (-nl + m)m + (-l - mn)l + (n^2 - 1)(-1) = 0 \]

Expanding this gives:

\[ -mnl + m^2 - l^2 - lmn - n^2 + 1 = 0 \]

Rearranging leads to:

\[ m^2 + n^2 + l^2 + 2mln - 1 = 0 \]

### Step 5: Final equation
Thus, the condition for \( l, m, n \) is:

\[ l^2 + m^2 + n^2 + 2mln = 1 \]

### Conclusion
The values of \( l, m, n \) must satisfy the equation \( l^2 + m^2 + n^2 + 2mln = 1 \).