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Question

# If the vectors a^i+^j+^k, ^i+b^j+^k and ^i+^j+c^k, (a≠b≠c≠1) are coplanar, then find the value of 11−a+11−b+11−c.

A
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B
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C
Not defined
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D
None of the above
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Solution

## The correct option is C 1Since ai+j+k,i+bj+k and i+j+ck=0 are coplanar, there exist scalar x,y,z (not all zero) such thatx(ai+j+k)+y(i+bj+k)+z(i+j+ck)=0⇒(ax+y+z)i+(x+by+z)j+(x+y+cz)k=0⇒ax+y+z=0,x+by+z=0,x+y+cz=0Since at least one of x,y,z is different from zero, the above system must have a non-zero solution.∴∣∣ ∣∣a111b111b∣∣ ∣∣=0⇒∣∣ ∣∣a111−2b−101−a0c−1∣∣ ∣∣=0⇒a(b−1)(c−1)−(1−a)(b−1)−(c−1)(1−a)=0Dividing by (1−a)(1−b)(1−c), we geta1−a+11−c+11−b=0⇒1−(1−a)1−a+11−c+11−b=0⇒11−a+11−b+11−c=1

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