1. Given quadratic and its roots

The roots $\alpha,\beta$ satisfy $x^2-\sqrt{2}x-\sqrt{3}=0.$ Hence each root satisfies $r^2=\sqrt{2}\,r+\sqrt{3}.$

Also, $P_n=\alpha^n-\beta^n.$

2. Recurrence relation for $P_n$

Since $\alpha^2=\sqrt{2}\alpha+\sqrt{3}$, multiplying by $\alpha^n$ gives $\alpha^{n+2}=\sqrt{2}\alpha^{n+1}+\sqrt{3}\alpha^n.$ Similarly, $\beta^{n+2}=\sqrt{2}\beta^{n+1}+\sqrt{3}\beta^n.$ Subtracting, $P_{n+2}=\sqrt{2}P_{n+1}+\sqrt{3}P_n.$ So, $P_{12}=\sqrt{2}P_{11}+\sqrt{3}P_{10},$ $P_{11}=\sqrt{2}P_{10}+\sqrt{3}P_9.$

3. Evaluate the given expression

We need to compute $E=(11\sqrt{3}-10\sqrt{2})P_{10}+(11\sqrt{2}+10)P_{11}-11P_{12}.$

Now use $P_{12}=\sqrt{2}P_{11}+\sqrt{3}P_{10}.$ Then \begin{align*} E&=(11\sqrt{3}-10\sqrt{2})P_{10}+(11\sqrt{2}+10)P_{11}-11(\sqrt{2}P_{11}+\sqrt{3}P_{10})\ &=(11\sqrt{3}-10\sqrt{2}-11\sqrt{3})P_{10}+\big((11\sqrt{2}+10)-11\sqrt{2}\big)P_{11}\ &=-10\sqrt{2},P_{10}+10P_{11}\ &=10(P_{11}-\sqrt{2}P_{10}). \end{align*}

Now use $P_{11}=\sqrt{2}P_{10}+\sqrt{3}P_9.$ So, $P_{11}-\sqrt{2}P_{10}=\sqrt{3}P_9.$

Therefore, $E=10\sqrt{3}P_9.$

4. Compare with options

Thus the expression equals $10\sqrt{3}P_9,$ which is Option A.

5. Comparison with stored answer

Stored correct answer: A

This matches our derived answer.