Plane passes through (2,1,-1) and (-1,3,4),

A(x-2) + B(y-1) + C(z + 1)=0 (1)

A(x + 1) + B(y-3) + C(z-4)=0 (2)

Subtracting (1) from (2),

A(x + 1-x + 2) + B(y-3-y + 1) + C(z-4-z-1)=0

3A-2B-5C=0 (3)

Now plane is perpendicular to x-2y + 4z=10

A-2B + 4C=0 (4)

Using (3) in (4)

2A-9C=0

Putting values in equation (1)

A(18(x-2) + 17(y-1) + 4(z + 1)=0

18x + 17y + 4z-36-17 + 4=0

18x + 17y + 4z-49=0

So, the required equation of plane is 18x + 17y + 4z-49=0

LHS=18(-1) + 17.3 + 4.4-49

=-18 + 51 + 16-49

=-2 + 2=0=RHS

In vector form normal of plane

LHS=18.3 + 17(-2) + 4.(-5)=54-34-20=0=RHS

Hence line is contained in plane.