\(\overrightarrow{AB}=\left(-1;1\right)\) nên pt AB có dạng:
\(1\left(x-2\right)+1\left(y-3\right)=0\Leftrightarrow x+y-5=0\)
Do I thuộc AB nên tọa độ có dạng: \(I\left(a;5-a\right)\Rightarrow\left\{{}\begin{matrix}\overrightarrow{IA}=\left(2-a;a-2\right)\\\overrightarrow{IB}=\left(1-a;a-1\right)\\\overrightarrow{IC}=\left(-1-a;a-10\right)\end{matrix}\right.\)
\(\Rightarrow\overrightarrow{IA}+3\overrightarrow{IB}+5\overrightarrow{IC}=\left(-9a;9a-55\right)\)
\(\Rightarrow\left|\overrightarrow{IA}+3\overrightarrow{IB}+5\overrightarrow{IC}\right|=\sqrt{\left(9a\right)^2+\left(55-9a\right)^2}\ge\sqrt{\dfrac{1}{2}\left(9a+55-9a\right)^2}=\dfrac{55}{\sqrt{2}}\)
Dấu "=" xảy ra khi \(9a=55-9a\Rightarrow a=\dfrac{55}{18}\Rightarrow I\left(\dfrac{55}{18};\dfrac{35}{18}\right)\)
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