Let P be a real number and |P| geq 2. If A, B | vedclass

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(C) Let $\vec{u} = \sqrt{P^2-4} \hat{i} + P \hat{j} + \sqrt{P^2+4} \hat{k}$ and $\vec{v} = 	an A \hat{i} + 	an B \hat{j} + 	an C \hat{k}$.By the do

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$12$

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(C) Let $\vec{u} = \sqrt{P^2-4} \hat{i} + P \hat{j} + \sqrt{P^2+4} \hat{k}$ and $\vec{v} = 	an A \hat{i} + 	an B \hat{j} + 	an C \hat{k}$.By the dot product definition,$\vec{u} \cdot \vec{v} = \sqrt{P^2-4} 	an A + P 	an B + \sqrt{P^2+4} 	an C = 6P$.We know that $\vec{u} \cdot \vec{v} = |\vec{u}| |\vec{v}| \cos 	heta$,where $	heta$ is the angle between $\vec{u}$ and $\vec{v}$.First,calculate the magnitude of $\vec{u}$:$|\vec{u}| = \sqrt{(\sqrt{P^2-4})^2 + P^2 + (\sqrt{P^2+4})^2} = \sqrt{P^2-4 + P^2 + P^2+4} = \sqrt{3P^2} = \sqrt{3} |P|$.Thus,$|\vec{u}| |\vec{v}| \cos 	heta = \sqrt{3} |P| \sqrt{	an^2 A + 	an^2 B + 	an^2 C} \cos 	heta = 6P$.Since $|P| \geq 2$,we can write $|P| = P$ (or consider the magnitude),so $\sqrt{	an^2 A + 	an^2 B + 	an^2 C} = \frac{6P}{\sqrt{3} P \cos 	heta} = 2\sqrt{3} \sec 	heta$.Squaring both sides:$	an^2 A + 	an^2 B + 	an^2 C = (2\sqrt{3})^2 \sec^2 	heta = 12 \sec^2 	heta$.Since $\sec^2 	heta \geq 1$,the minimum value is $12(1) = 12$.

Let $P$ be a real number and $|P| \geq 2$. If $A, B, C$ are variable angles such that $(\sqrt{P^2-4}) \tan A + P \tan B + (\sqrt{P^2+4}) \tan C = 6P$,then the minimum value of $\tan^2 A + \tan^2 B + \tan^2 C$ is:

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