If $$\left\langle l, m, n \right\rangle,$$are the direction cosines of normal to the plane $$\quad 2x - 3y + 6z + 4 = 0, \quad$$ then what is the value of $$49 \left( 7l^2 + m^2 - n^2 \right)$$​​?

Correct option is B

​$$\vec{n} = (2, -3, 6) \\\text{DC's} = \left( \frac{2}{7}, \frac{-3}{7}, \frac{6}{7} \right) \\\text{then,} \\\Rightarrow 49\left( 7l^2 + m^2 - n^2 \right) \\\Rightarrow 49\left( 7\left( \frac{4}{49} \right) + \frac{9}{49} - \frac{36}{49} \right) \\\Rightarrow 28 + 9 - 36 \\\Rightarrow 37 - 36 \\\Rightarrow 1$$​​