\(\begin{array}{l}
P = 3 - 4x - {x^2} = 7 - \left( {{x^2} - 4x + 4} \right) = 7 - {\left( {x - 2} \right)^2}\\
Do\,\,{\left( {x - 2} \right)^2} \ge 0 \Rightarrow 7 - {\left( {x - 2} \right)^2} \le 7\\
\Rightarrow {P_{\max }} = 7 \Leftrightarrow x - 2 = 0 \Leftrightarrow x = 2\\
Q = 2x - 2 - 3{x^2} = - \left( {3{x^2} - 2x} \right) - 2\\
= - 3\left( {{x^2} - \frac{2}{3}x} \right) - 2\\
= - 3\left( {{x^2} - 2.x.\frac{1}{3} + \frac{1}{9}} \right) - 2 + \frac{1}{3}\\
= - \frac{5}{3} - 3{\left( {x - \frac{1}{3}} \right)^2} \le - \frac{5}{3} \Rightarrow {Q_{\max }} = - \frac{5}{3} \Leftrightarrow x = \frac{1}{3}\\
R = 2 - {x^2} - {y^2} - 2x - 2y\\
= - \left( {{x^2} + 2x + 1} \right) - \left( {{y^2} + 2y + 1} \right) + 4\\
= 4 - {\left( {x + 1} \right)^2} - {\left( {y + 1} \right)^2} \le 4\\
{R_{\max }} = 4 \Leftrightarrow x = - 1;y = - 1
\end{array}\)
