Question

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Answer 1

$1)\; \; x_0=1\; \; \to \; \; \left \{ {{2-y+z+1=0} \atop {1+3y+2z-5=0}} \right. \; \left \{ {{-y+z=-3} \atop {3y+2z=4}} \right. \; \left \{ {{y=z+3} \atop {5z=-5}} \right. \; \left \{ {{y=2} \atop {z=-1}} \right. \\\\M_0(1,2,-1)\\\\\vec{n}=\left|\begin{array}{ccc}i&j&k\\2&-1&1\\1&3&2\end{array}\right|=-5i-3j+7k\\\\-5(x-1)-3(y-2)+7(z+1)=0\\\\\underline {4x+3y-7z-18=0}$

$2)\; \; M_0(1,-1,3)\; \; ,\; \; \pi :\; x+2y-z-2=0\; ,\; \; l\perp \pi \\\\\vec{n}=\vec{s}=(1,2,-1)\\\\l:\frac{x-1}{1}=\frac{y+1}{2}=\frac{z-3}{-1}\\\\l:\; \left\{\begin{array}{ccc}x=t+1\\y=2t-1\\z=-t+3\end{array}\right \\\\l\cap \pi =M_1\; \; \to \; \; (t+1)+2(2t-1)-(-t+3)-2=0\; ,\\\\6t-4=0\; ,\; \; t=\frac{2}{3}\\\\M_1:\; \; \left\{\begin{array}{ccc}x_1=\frac{2}{3}+1=\frac{5}{3}\\y_1=2\cdot \frac{2}{3}-1=\frac{1}{3}\\z_1=-\frac{2}{3}+3=\frac{7}{3}\end{array}\right\\\\\\\underline {M_1(\, \frac{5}{3}\, ,\, \frac{1}{3}\, ,\, \frac{7}{3}\, )}$