To solve this problem, we need to find the values of the direction cosines of a normal to the given plane equation and then evaluate the expression provided.

The equation of the plane is \(2x - 3y + 6z + 4 = 0\). The normal vector to this plane is given by the coefficients of \(x\), \(y\), and \(z\) from the plane equation, which is \(2\mathbf{i} - 3\mathbf{j} + 6\mathbf{k}\).

To find the direction cosines \(l\), \(m\), and \(n\), we use the components of the normal vector:

• The magnitude of the normal vector is \(\sqrt{2^2 + (-3)^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7\).
• The direction cosines are calculated by dividing each component of the normal vector by its magnitude:
  • \(l = \frac{2}{7}\)
  • \(m = \frac{-3}{7}\)
  • \(n = \frac{6}{7}\)

Next, we need to evaluate the expression \(49(7l^2 + m^2 - n^2)\).

• Calculate \(l^2 = \left(\frac{2}{7}\right)^2 = \frac{4}{49}\)
• Calculate \(m^2 = \left(\frac{-3}{7}\right)^2 = \frac{9}{49}\)
• Calculate \(n^2 = \left(\frac{6}{7}\right)^2 = \frac{36}{49}\)
• Substitute back into the expression:
  • \(7l^2 = 7 \times \frac{4}{49} = \frac{28}{49}\)
  • The expression becomes \(49\left(\frac{28}{49} + \frac{9}{49} - \frac{36}{49}\right)\)

Simplify inside the parentheses:

• \(\frac{28}{49} + \frac{9}{49} - \frac{36}{49} = \frac{28 + 9 - 36}{49} = \frac{1}{49}\)

Finally, multiply by 49:

• \(49 \times \frac{1}{49} = 1\)

Thus, the value of \(49(7l^2 + m^2 - n^2)\) is 1.