Find $|2a+3b+2\sqrt{3}(a\times b)|$ where a,b are perpendicular unit vectors.

My attempt:

$(|2a+3b+2\sqrt{(}3)(a\times b)|{)}^{2}$

$(a.b=0,|a|=|b|=1),\hat{n}$ is normal vector to a and b.

$4+9+12+8\sqrt{3}(\pm 1)+12\sqrt{3}(\pm 1)$ $25\pm 8\sqrt{3}\pm 12\sqrt{3}|$ 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.

My attempt:

$(|2a+3b+2\sqrt{(}3)(a\times b)|{)}^{2}$

$(a.b=0,|a|=|b|=1),\hat{n}$ is normal vector to a and b.

$4+9+12+8\sqrt{3}(\pm 1)+12\sqrt{3}(\pm 1)$ $25\pm 8\sqrt{3}\pm 12\sqrt{3}|$ 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.