Question 941323
Let {{{u}}} and {{{v}}} be two vectors in the plane. 
{{{u}}}=({{{x[1]}}},{{{y[1]}}},{{{z[1]}}})
{{{v}}}=({{{x[2]}}},{{{y[2]}}},{{{z[2]}}})
Since they're in the plane,
1.{{{2x[1]+4y[1]+6z[1]=12}}}
2.{{{2x[2]+4y[2]+6z[2]=12}}}
If you subtract eq. 1 from eq. 2,
3.{{{2(x[2]-x[1])+4(y[2]-y[1])+6(z[2]-z[1])=0}}}
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If they are perpendicular to the normal vector, then the dot product of each vector with the normal will be zero.
{{{u*n=0}}}
({{{x[1]}}},{{{y[1]}}},{{{z[1]}}})*({{{2}}},{{{4}}},{{{6}}})={{{0}}}
4.{{{2x[1]+4y[1]+6z[1]=0}}}
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{{{v*n=0}}}
({{{x[2]}}},{{{y[2]}}},{{{z[2]}}})*({{{2}}},{{{4}}},{{{6}}})={{{0}}}
5.{{{2x[2]+4y[2]+6z[2]=0}}}
So then if I subtract the two dot products,
6.{{{2(x[2]-x[1])+4(y[2]-y[1])+6(z[2]-z[1])=0}}}
Comparing eq. 6 to eq. 3, you recognize they are identical.