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Question

The position vector of the points A,B,C,D are 3^i2^j^k,2^i+3^j4^k,^i+^j+2^k,4^i+5^j+γ^k respectively. If the points A,B,C,D lie on a plane the value of γ is

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Solution

If the points are coplanar then their position vectors a,b,c,d must satisfy
[ba,ca,da]=0
Let a=3^i2^j^k,b=2^i+3^j4^k,c=^i+^j+2^k,d=4^i+5^j+γ^k
ba=(2^i+3^j4^k)(3^i2^j^k)
=^i+5^j3^k
ca=(ba)(3^i2^j^k)
=4^i+3^j+3^k
da=(4^i+5^j+γ^k)(3^i2^j^k)
=^i+7^j+(1+γ)^k
Now [ba,ca,da]
=∣ ∣15343317γ+1∣ ∣
=1(3γ+321)5(4γ43)3(283)
=1(3γ18)5(4γ7)3(31)
146+17γ=0 on simplification
γ=14617

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