Correct option is (B) (1/√2 , 1/√6, 1/√3)
AAT = I
⇒ \(\begin{bmatrix}0&2m&n\\l&m&-n\\l&-m&n\end{bmatrix}\)\(\begin{bmatrix}0&l&l\\2m&m&-n\\n&-n&n\end{bmatrix}\) = \(\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}\)
⇒ \(\begin{bmatrix}4m^2+n^2&2m^2-n^2&-2m^2+n^2\\2m^2-n^2&l^2+m^2+n^2&l^2-m^2-n^2\\-2m^2+n^2&l^2-m^2-n^2&l^2+m^2+n^2\end{bmatrix}\) = \(\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}\)
⇒ 4m2 + n2 = 1
2m2 - n2 = 0
⇒ 6m2 = 1 ⇒ m2 = 1/6 ⇒ m = \(\pm\frac1{\sqrt6}\)
\(\therefore\) n2 = 2m2 = 2/6 = 1/3 ⇒ n = \(\pm\frac1{\sqrt3}\)
Also, l2 + m2 + n2 = 1
⇒ l2 = 1 - 1/6 - 1/3 = 3/6 = 1/2
⇒ l = \(\pm\frac1{\sqrt2}\)
Hence, l, m, n = (1/√2 , 1/√6, 1/√3)