BT dbts

See first we need to find the eqⁿ of the normal to the plane P₁ by finding the cross product of ( 2j + 3k ) and (4j - 3k ) which comes out to be -18i
then similarly eqⁿ of the normal to the plane plane P₂ = 3i - 3j - 3k
then to find the eqⁿ of line passing through the intersection of the abv plane we need to find the cross product P₁ P₂..
which comes out to be 54j - 54k..
now find dot product 54j - 54k with 2i + 2j - k
cosθ = 1/√2

so θ = π/4...

Ans A

thanks for pointing out the mistake ..che
edited that part
the cross product of ( 2j + 3k ) and (4j - 3k ) gives u the normal vector to the plane P ₁

@govind

the cross product of ( 2j + 3k ) and (4j - 3k ) gives u the normal vector to the plane P ₁not the eq. of teh plane P₁ !

now cross product of the 2 normal vectors to the 2 respective planes gives u the vector which is paralle to the line of intersection of planes