Помогите пожалуйста 5.23

Question 1

Question 2

Решите задачу:

$5.23.\; \; \; l:\left \{ {{2x-y+z-3=0} \atop {4y-z=0\qquad }} \right.\\\\\\a)\; \; \vec{s}=\vec{n}_1\times \vec{n}_2\\\\\vec{s}=\left|\begin{array}{ccc}i&j&k\\2&-1&1\\0&4&-1\end{array}\right|=i(1-4)-j(-2-0)+k(8-0)\\\\\\\vec{s}=-3\vec{i}+2\vec{j}+8\vec{k}\\\\x=0:\; \; \left \{ {{-y+z=3} \atop {4y-z=0}} \right.\; \oplus \; \left \{ {{z=3+y} \atop {3y=3}} \right. \; \left \{ {{z=4} \atop {y=1}} \right. \; \; \Rightarrow \; \; \; B(0,1,4)\in l\\\\\boxed {l:\; \; \frac{x}{-3}=\frac{y-1}{2}=\frac{z-4}{8}}$

$b)\; \; A\notin l\; ,\; \; l\parallel l_1\; ,\; A\in l_1\\\\A(1,-1,2)\notin l\; ,\; t.k.\; \; \left \{ {{2\cdot 1-(-1)+2-3=2\ne 0} \atop {4\cdot (-1)-2=-6\ne 0}} \right. \\\\l\parallel l_1\; \; \Rightarrow \; \; \vec{s}=\vec{s}_1=(-3,2,8)\\\\\boxed {l_1:\; \; \frac{x-1}{-3}=\frac{y+1}{2}=\frac{z-2}{8}}\\\\c)\; \; l\in \alpha \; ,\; \; l_1\in \alpha \; ,\; \; A,B\in \alpha \; ,\; \; \alpha \; -\; ?\\\\\overline {AB}\in \alpha \; ,\; \; \overline {AB}=(0-1,\, 1+1,\, 4-2)=(-1,2,2)$

$\overline {AB}\times \vec{s}=\left|\begin{array}{ccc}i&j&k\\-1&2&2\\-3&2&8\end{array}\right|=i(16-4)-j(-8+6)+k(-2+6)=\\\\\\=12i+2j+4k\; \; \; \Rightarrow \; \; \; \; \vec{n}=\frac{1}{2}\cdot [\, \overline {AB}\times \vec{s}\, ]=(6,1,2)\\\\\alpha :\; \; 6\cdot (x-0)+1\cdot (y-1)+2\cdot (z-4)=0\\\\\boxed {\alpha :\; \; \; 6x+y+2z-9=0}$