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

Explanation

The direction ratios of the normal are \langle 2, -3, 6 \rangle. Its magnitude is \sqrt{2^2 + (-3)^2 + 6^2} = \sqrt{4+9+36} = 7. The direction cosines are l = \frac{2}{7}, m = -\frac{3}{7}, n = \frac{6}{7}. Substituting these values into 49(7l^2+m^2-n^2): 49(7(\frac{4}{49}) + \frac{9}{49} - \frac{36}{49}) = 49(\frac{28+9-36}{49}) = 28+9-36 = 1.

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