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\left(\sqrt{2}\right)^{2}-2\sqrt{2}\sqrt{3}+\left(\sqrt{3}\right)^{2}-2\times 3\sqrt{\frac{1}{3}}\sqrt{12}
Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(\sqrt{2}-\sqrt{3}\right)^{2}.
2-2\sqrt{2}\sqrt{3}+\left(\sqrt{3}\right)^{2}-2\times 3\sqrt{\frac{1}{3}}\sqrt{12}
The square of \sqrt{2} is 2.
2-2\sqrt{6}+\left(\sqrt{3}\right)^{2}-2\times 3\sqrt{\frac{1}{3}}\sqrt{12}
To multiply \sqrt{2} and \sqrt{3}, multiply the numbers under the square root.
2-2\sqrt{6}+3-2\times 3\sqrt{\frac{1}{3}}\sqrt{12}
The square of \sqrt{3} is 3.
5-2\sqrt{6}-2\times 3\sqrt{\frac{1}{3}}\sqrt{12}
Add 2 and 3 to get 5.
5-2\sqrt{6}-6\sqrt{\frac{1}{3}}\sqrt{12}
Multiply 2 and 3 to get 6.
5-2\sqrt{6}-6\times \frac{\sqrt{1}}{\sqrt{3}}\sqrt{12}
Rewrite the square root of the division \sqrt{\frac{1}{3}} as the division of square roots \frac{\sqrt{1}}{\sqrt{3}}.
5-2\sqrt{6}-6\times \frac{1}{\sqrt{3}}\sqrt{12}
Calculate the square root of 1 and get 1.
5-2\sqrt{6}-6\times \frac{\sqrt{3}}{\left(\sqrt{3}\right)^{2}}\sqrt{12}
Rationalize the denominator of \frac{1}{\sqrt{3}} by multiplying numerator and denominator by \sqrt{3}.
5-2\sqrt{6}-6\times \frac{\sqrt{3}}{3}\sqrt{12}
The square of \sqrt{3} is 3.
5-2\sqrt{6}-6\times \frac{\sqrt{3}}{3}\times 2\sqrt{3}
Factor 12=2^{2}\times 3. Rewrite the square root of the product \sqrt{2^{2}\times 3} as the product of square roots \sqrt{2^{2}}\sqrt{3}. Take the square root of 2^{2}.
5-2\sqrt{6}-12\times \frac{\sqrt{3}}{3}\sqrt{3}
Multiply 6 and 2 to get 12.
5-2\sqrt{6}-4\sqrt{3}\sqrt{3}
Cancel out 3, the greatest common factor in 12 and 3.
5-2\sqrt{6}-4\times 3
Multiply \sqrt{3} and \sqrt{3} to get 3.
5-2\sqrt{6}-12
Multiply 4 and 3 to get 12.
-7-2\sqrt{6}
Subtract 12 from 5 to get -7.