Respuesta :
The solution for the system of equaiton is (2,1,0).
Take the general form of the equation ax+by+cz+d=0.
For the first equation take the coefficient a and b as 1 then the substituting the known values,
[tex]\begin{gathered} ax+by+cz+d=0 \\ 1\times x+1\times y+cz+d=0 \\ x+y+cz+d=0 \\ 2+1+c(0)+d=0 \\ 3+d=0 \\ d=-3 \end{gathered}[/tex]As the value of d is independent of value of c we can choose any value of c arbitarily.
[tex]x+y+3z-3=0\ldots..\text{.}\mathrm{}(1)[/tex]For the second equation choose a and b as 0 and 1 respectively. Now substituting the known values,
[tex]\begin{gathered} ax+by+cz+d=0 \\ 0\times x+1\times y+cz+d=0 \\ y+cz+d=0 \\ 1+c\times0+d=0 \\ d=-1 \end{gathered}[/tex]As the value of d is independent of value of c we can choose any value of c arbitarily.
[tex]y+4z-1=0\ldots\ldots\text{.}(2)[/tex]For the third equation take the coefficients a and b as 1 and 0 respectively,
[tex]\begin{gathered} ax+by+cz+d=0 \\ 1\times x+0\times y+cz+d=0 \\ x+cz+d=0 \\ 2+c\times0+d=0 \\ d=-2 \end{gathered}[/tex]As the value of d is independent of value of c we can choose any value of c arbitarily.
[tex]x+5z-2=0\ldots\ldots.(3)_{}[/tex]