a) \(\vec{n}=(1,1,1)\) where \(\vec{n}\) is the normal to the plane that passes through (1,0,0)

\(\therefore\) Using the equation of the plane we get:

\(1(x-1)+1(y-0)+1(z-0)=0\)

\(x+y+z-1=0\)

b)\(\vec{n}=(1,2,3)\) where \(\vec{n}\) is the normal to the plane that passes through (1,1,1)

\(\therefore\) Using the equation of the plane we get:

\(1(x-1)+2(y-1)+3(z-1)=0\)

\(x+2y+3z-6=0\)

c) \(\vec{n}=\text{direction of line}=(5,0,2)\) where \(\vec{n}\) is the normal to the plane that passes through (5,-1,0)

\(\therefore\) Using the equation of the plane we get:

\(5(x-5)+0(y+1)+2(z-0)=0\)

\(5x+2z-25=0\)

d) \(\vec{n}=\text{direction of line}=(-1,-2,3)\) where \(\vec{n}\) is the normal to the plane that passes through (2,4,-1)

\(\therefore\) Using the equation of the plane we get:

\(-1(x-2)-2(y-4)+3(z+1)=0\)

\(-x-2y+3z+13=0\)

\(\therefore\) Using the equation of the plane we get:

\(1(x-1)+1(y-0)+1(z-0)=0\)

\(x+y+z-1=0\)

b)\(\vec{n}=(1,2,3)\) where \(\vec{n}\) is the normal to the plane that passes through (1,1,1)

\(\therefore\) Using the equation of the plane we get:

\(1(x-1)+2(y-1)+3(z-1)=0\)

\(x+2y+3z-6=0\)

c) \(\vec{n}=\text{direction of line}=(5,0,2)\) where \(\vec{n}\) is the normal to the plane that passes through (5,-1,0)

\(\therefore\) Using the equation of the plane we get:

\(5(x-5)+0(y+1)+2(z-0)=0\)

\(5x+2z-25=0\)

d) \(\vec{n}=\text{direction of line}=(-1,-2,3)\) where \(\vec{n}\) is the normal to the plane that passes through (2,4,-1)

\(\therefore\) Using the equation of the plane we get:

\(-1(x-2)-2(y-4)+3(z+1)=0\)

\(-x-2y+3z+13=0\)