<s>[INST] Find $|2a + 3b + 2 \sqrt{3}(a \times b)|$ where $a,b$ are perpendicular unit vectors. Find $|2a + 3b + 2 \sqrt{3}(a \times b)|$ where &a,b$ are perpendicular unit vectors.
My attempt$:$
$(|2a+3b+2\sqrt(3)(a \times b)|)^{2}$ = $ 4a^{2} + 9b^{2} + 12 + 8\sqrt{3} a.\hat{n} +12\sqrt{3}b.\hat{n}$
$( a.b = 0, |a| = |b| = 1), \hat{n}$ is normal vector to $a$ and $b$.
$4+9+12+8\sqrt{3} (±1) + 12\sqrt{3}(±1)$
$25 ± 8\sqrt{3} ± 12\sqrt{3}|$
I have to eliminate one case $ 25 - 20\sqrt{3}$ and then my answer should be $(25+20(3)^{1/2})^{1/2}$ or $(25+ 4(3)^{1/2})^{1/2}$ or $(25-4(3)^{1/2})^{1/2}$
Is this right way to solve the problem. I am given only one place to anwer and I have found three answers. 
 [/INST] $\{a,b,a\times b\}$ is an orthogonal set in $\mathbb{R}^3$ so the Pythagorean theorem gives:
$$|2a + 3b + 2\sqrt{3}(a\times b)|^2 = |2a|^2 + |3b|^2 + |2\sqrt{3}(a\times b)|^2 = 4 + 9 + 12 = 25$$
Hence $|2a + 3b + 2\sqrt{3}(a\times b)| = 5$.
</s>