Question 940706
So the line is then,
({{{4-2t}}},{{{2-t}}},{{{1+0t}}})=({{{4-2t}}},{{{2-t}}},{{{1}}})
Substitute this into the plane equation and solve for {{{t}}}.
{{{2x-y+3z=6}}}
{{{2(4-2t)-(2-t)+3(1)=6}}}
{{{8-4t-2+t+3=6}}}
{{{9-3t=6}}}
{{{-3t=-3}}}
{{{t=1}}}
So then,
({{{4-2}}},{{{2-1}}},{{{1}}})=({{{2}}},{{{1}}},{{{1}}}) is the intersection point.