Find the equation of a plane that passes through the origin and contains a,b, and c?

2 vectors in R^3 will already define a plane. What you need to check is whether the third vector is on the plane or not. To do so, first, choose two out of the three vectors and cross them

<2,1,3> x <1,1,-1> = <-4,5,1>

if the the third vector is on the plane, then it is perpendicular to <-4,5,1>. Two vectors are perpendicular if their dot product is 0

so,

<-4,5,1> dot <1,-1,9> = -4(1) - 5(1) + 1(9) = -9 - 9 = 0

so the third vector, indeed, is on the plane. Which is obvious, since the question already states that they all are on the same plane; but generally, you have to check.

the equation of a plane given a normal vector <a,b,c> and passes through point (xo, yo, zo) is:

a(x-xo) + b(y-yo) + c(z-zo) = 0

in this case <a,b,c> = <-4,5,1> and (xo, yo, zo) = (0,0,0). So

-4x + 5y + z = 0

I hope this helps