Question

 

Answer 1

Lời giải:
$B,C\in (P)$ nên: \(\left\{\begin{matrix} a+2c=3\\ -a-b=3\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} c=\frac{3-a}{2}\\ b=-3-a\end{matrix}\right.\)

\(d(A,(P))=\frac{|2a+5b+3c-3|}{\sqrt{a^2+b^2+c^2}}=\frac{3|a+3|}{\sqrt{a^2+2a+5}}\)

$a^2+2a+5=(a+1)^2+4=(a+1)^2+2^2\geq \frac{(a+1+2)^2}{2}=\frac{(a+3)^2}{2}$ theo BĐT Cô si dạng $x^2+y^2\geq \frac{(x+y)^2}{2}$

$\Rightarrow \sqrt{a^2+2a+5}\geq \frac{|a+3|}{\sqrt{2}}$

$\Rightarrow d(A,(P))\leq \sqrt{2}$. Vậy $d(A,(P))_{\max}=3\sqrt{2}$ khi $a+1=2\Leftrightarrow a=1$

$b=-4; c=1$

Khi đó: $T=a+b+c=1+(-4)+1=-2$

Đáp án A.