Câu hỏi:

\(\int f(x) d x=\int\left(x^{2}+4 x+2\right) d x=\frac{x^{3}}{3}+2 x^{2}+2 x+C\)

\(\int_{0}^{3}\left[2 f(x)+3^{-x}\right] d x=2 \int_{0}^{3} f(x) d x+\int_{0}^{3} 3^{-x} d x=\left.2\left(x^{2}+4 x+2\right)\right|_{0} ^{3}+\int_{0}^{3}\left(\frac{1}{3}\right)^{x} d x=66+\left.\frac{\left(\frac{1}{3}\right)^{x}}{\ln \frac{1}{3}}\right|_{0} ^{3}\)

\(=66-\frac{26}{27 \ln \frac{1}{3}}=66+\frac{26}{27 \ln 3}\)

Do đó \(a=66, c=26, d=27\). Vậy \(T=a-c-d=66-26-27=13\).

\(\overrightarrow{C B}=(3 ;-2 ; 1)\)

Gọi \(I\) là trung điểm \(BC\) , khi đó \(I\left(\frac{1}{2} ; 2 ; \frac{3}{2}\right)\).

Ta có: \(R=\frac{B C}{2}=\frac{\sqrt{3^{2}+(-2)^{2}+1^{2}}}{2}=\frac{\sqrt{14}}{2}\).

Khi đó phương trình mặt cầu là:

\((S):\left(x-\frac{1}{2}\right)^{2}+(y-2)^{2}+\left(z-\frac{3}{2}\right)^{2}=\frac{7}{2}\).

Gọi \(M(x ; y ; z)\)

Ta có: \(\overrightarrow{M A}+2 \overrightarrow{M B}-\overrightarrow{M C}=\overrightarrow{0} \Rightarrow 2 \overrightarrow{M B}+\overrightarrow{C A}=\overrightarrow{0} \Rightarrow 2 \overrightarrow{M B}=\overrightarrow{A C} \Rightarrow \overrightarrow{M B}=\frac{1}{2} \overrightarrow{A C}\)

Suy ra \(\left\{\begin{array}{l}2-x=\frac{1}{2}(-1-1) \\ 1-y=\frac{1}{2}(3-2) \\ 2-z=\frac{1}{2}(1-0)\end{array} \Leftrightarrow\left\{\begin{array}{l}x=3 \\ y=\frac{1}{2} \\ z=\frac{3}{2}\end{array} \Rightarrow M\left(3 ; \frac{1}{2} ; \frac{3}{2}\right)\right.\right.\).

\(\overrightarrow{A B}=(1 ;-1 ; 2), \overrightarrow{A C}=(-2 ; 1 ; 1), \overrightarrow{B C}=(-3 ; 2 ;-1)\)

Suy ra \(A B=A C=\sqrt{6}, B C=\sqrt{14}\)

\([\overrightarrow{A B}, \overrightarrow{A C}]=(-3 ;-5 ;-1) \Rightarrow S_{\triangle A B C}=\frac{1}{2}|[\overrightarrow{A B}, \overrightarrow{A C}]|=\frac{\sqrt{35}}{2}\)

Mà ta có: \(S_{\triangle A B C}=\frac{A B \cdot B C \cdot C A}{4 \cdot R_{\triangle A B C}} \Rightarrow R_{\triangle A B C}=\frac{A B \cdot B C \cdot C A}{4 \cdot S_{\triangle A B C}}=\frac{3 \sqrt{10}}{5}\).