Trong không gian Oxyz cho hình chóp S.ABC

* Hướng dẫn giải

Chọn A

Ta có $\overrightarrow{SA\text{'}}=\frac{1}{3}\overrightarrow{SA};\overrightarrow{SB\text{'}}=\frac{1}{4}\overrightarrow{SB};\overrightarrow{SC\text{'}}=\frac{1}{5}\overrightarrow{SC};\overrightarrow{SG\text{'}}=k\overrightarrow{SG}.$

Bốn điểm $A\text{'},B\text{'},C\text{'},G\text{'}$ đồng phẳng nên với mọi điểm S ta có $\overrightarrow{SG\text{'}}=x\overrightarrow{SA\text{'}}+y\overrightarrow{SB\text{'}}+z\overrightarrow{SC\text{'}}\left(1\right)$ với $x+y+z=1.$

$\left(1\right)\iff k\overrightarrow{SG}=\frac{x}{3}\overrightarrow{SA}+\frac{y}{4}\overrightarrow{SB}+\frac{z}{5}\overrightarrow{SC},$ mặt khác $\overrightarrow{SG}=\frac{1}{3}\left(\overrightarrow{SA}+\overrightarrow{SB}+\overrightarrow{SC}\right).$ Vì $\overrightarrow{SA},\overrightarrow{SB},\overrightarrow{SC}$ không đồng phẳng nên $\left\{\begin{array}{l}\frac{k}{3}=\frac{x}{3}\\ \frac{k}{3}=\frac{y}{4}\\ \frac{k}{3}=\frac{z}{5}\end{array}\right.\iff \left\{\begin{array}{l}x=k\\ y=\frac{4}{3}k\\ z=\frac{5}{3}k\end{array}\right.;x+y+z=1\iff k+\frac{4}{3}k+\frac{5}{3}k=1\iff k=\frac{1}{4}.$

Vậy $\overrightarrow{SG\text{'}}=\frac{1}{4}\overrightarrow{SG}=\frac{1}{4}\left(-3;-1;-1\right)\iff \left\{\begin{array}{l}a-2=\frac{-3}{4}\\ b-3=\frac{-1}{4}\\ c-1=\frac{-1}{4}\end{array}\right.\Rightarrow a+b+c=6-\frac{5}{4}=\frac{19}{4}.$