Answer :

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.


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