The equation of the plane through the intersection of the planes x+2y+3z-4=0 and 4x+3y+22+ 1 = 0 and passing through the origin is
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a
17x+14y+11z=0
b
7x+4y+z=0
c
x+14y+11z=0
d
17x+y+z=0
answer is A.
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Detailed Solution
Any plane through the given planes is x+2y+3z−4+λ(4x+3y+2z+1)=0It passes through (0, 0, 0). Therefore, -4+λ=0 or, λ=4Therefore, the required plane is x + 2y + 3z + 4(4x + 3y + 2z) = 0 or 17x + 14y + 11z=0.