Answer:
The equation of plane is x-y+z=12.
Step-by-step explanation:
It is given that the plane passing through the point (0,0,6) perpendicular to x=1-t y=2+t z=4-2t.
The given line is perpendicular to required plane, so coefficients of t represents the normal vector.
Normal vector is
[tex]\overrightarrow {n}=<-1,1,-2>[/tex]
If a plane passes through [tex](x_1,y_1,z_1)[/tex] and having normal vector [tex]\overrightarrow {n}=<a,b,c>[/tex], then the equation of plane is
[tex]a(x-x_1)+b(y-y_1)+c(z-z_1)=0[/tex]
[tex]-1(x-0)+1(y-0)+(-2)(z-6)=0[/tex]
[tex]-x+y-2z+12=0[/tex]
[tex]-x+y-2z=-12[/tex]
Multiply both sides by -1.
[tex]x-y+2z=12[/tex]
Therefore, the equation of plane is x-y+z=12.
