The question provides two equations involving the direction cosines \( l, m, \) and \( n \) of a line:

• \(l + 2m + n = 0\)
• \(2l - 2m + 3n = 0\)

We need to find the value of \(l^2 + m^2 - n^2\).

Since \( l, m, \) and \( n \) are direction cosines, they must satisfy the relation:

• \(l^2 + m^2 + n^2 = 1 \quad (1)\)

Let’s solve the given system of linear equations to find \( l, m, \) and \( n \):

From the first equation:

• \(n = -l - 2m \quad (2)\)

Substitute equation (2) into the second equation:

• \(2l - 2m + 3(-l - 2m) = 0\)
• \(2l - 2m - 3l - 6m = 0\)
• \(-l - 8m = 0 \quad \Rightarrow \quad l = -8m \quad (3)\)

Substitute equation (3) back into equation (2):

• \(n = -(-8m) - 2m\)
• \(n = 8m - 2m = 6m \quad (4)\)

Now we have \( l = -8m \), \( n = 6m \). Substitute these into equation (1):

• \((-8m)^2 + m^2 + (6m)^2 = 1\)
• \(64m^2 + m^2 + 36m^2 = 1\)
• \(101m^2 = 1\)
• \(m^2 = \frac{1}{101}\)

Substitute back to find \( l^2 \) and \( n^2 \):

• \(l^2 = (-8m)^2 = 64m^2 = 64 \cdot \frac{1}{101} = \frac{64}{101}\)
• \(n^2 = (6m)^2 = 36m^2 = 36 \cdot \frac{1}{101} = \frac{36}{101}\)

Finally, calculate \( l^2 + m^2 - n^2 \):

• \(l^2 + m^2 - n^2 = \frac{64}{101} + \frac{1}{101} - \frac{36}{101}\)
• \(= \frac{64 + 1 - 36}{101}\)
• \(= \frac{29}{101}\)

Thus, the value of \( l^2 + m^2 - n^2 \) is \(\frac{29}{101}\).