Question

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Answer 1

$A_1(-1,-2,1)\; ,\; A_2(-2,-2,5)\; ,\; \; A_3(-3,-1,1)\; ,\; \; A_4(-1,0,3)\\\\1)\; \; A_1A_2\, :\; \; \frac{x+1}{-2+1}=\frac{y+2}{-2+2}=\frac{z-1}{5-1}\; \; \to \; \; \; \frac{x+1}{-1}=\frac{y+2}{[0]}=\frac{z-1}{4}\; \; \to \\\\\left\{\begin{array}{lll}x=-t-1\\y=-2\\z=4t+1\end{array}\right\\\\A_1A_3\, :\; \; \frac{x+1}{-3+1}=\frac{y+2}{-1+2}=\frac{z-1}{1-1}\; \; \to \; \; \; \frac{x+1}{-2}=\frac{y+2}{1}=\frac{z-1}{[\, 0]}\; \; \; \to \\\\\left\{\begin{array}{lll}x=-2t-1\\y=t-2\\z=1\end{array}\right$

$2)\; \; x_{M}=\frac{-1-2}{2}=-\frac{3}{2}\; ,\; \; y_{M}=\frac{-2-2}{2}=-2\; \; ,\; \; z_{M}=\frac{1+5}{2}=3\\\\\overline {A_3M}=(-\frac{3}{2}+3\; ,\; -2+1\, ,\, 3-1)=(\frac{3}{2}\, ,\, -1\, ,\, 2)\; \; \Rightarrow \; \; \\\\\lambda \cdot \overline {A_3M}\parallel \overline {A_3M}\; ,\; \; \lambda =2\; \; \; \Rightarrow \; \; \; \lambda \cdot \overline {A_3M}=(3,-2,4)\\\\A_3M\, :\; \; \frac{x+3}{3}=\frac{y+1}{-2}=\frac{z-1}{4}\; \; \; ili\; \; \; \left\{\begin{array}{lll}x=3t-3\\y=-2t-1\\z=4t+1\end{array}\right$