The equation of the plane passes through the point (i+2j−k)=(1,2,−1) having direction ratios normal to a,b,c is, a(x−1)+b(y−2)+c(z+1)=0..........(I) The plane in equation (I) is perpendicular to line of intersection of planes r⋅(3i−j+k)=1 and r⋅(i+4j−2k)=2so 3a−b+c=0 a+4b−2c=0 This implies,
a
2−4
=
−b
−6−1
=
c
12+1
a
2
=
b
−7
=
c
−13
So, equation of required plane is 2(x−1)−7(y−2)−13(z+1)=0 2x−7y−13z−2+14−13=0 2x−7y−13z=1 r⋅(2i−7j−13k)=1