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Q.

If ax12+by12+cz12=ax22+by22+cz22=ax32+by32+cz32=λ and ax1x2+by1y2+cz1z2=ax2x3+by2y3+cz2z3=ax3x1+by3y1+cz3z1=μ and x1    y1    z1x2    y2    z2x3    y3    z3=(λμ)βλ+θμabcthen the value of θ+β is

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a

2

b

1

c

3

d

4

answer is A.

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Detailed Solution

LHS=x1y1z1x2y2z2x3y3z3×x1y1z1x2y2z2x3y3z3=1abcax1by1cz1ax2by2cz2ax3by3cz3×x1y1z1x2y2z2x3y3z3=ax12+by12+cz12ax1x2+by1y2+cz1z2ax3x1+by3y1a1x2+by1y2+cz1z2ax22+by22+cz22ax2x3+by2y3+cz2z3a3x1+by3y1+cz3z1ax2x3+by2y3+cz2z3ax32+by32+cz32=1abcλμμμλμμμλC1C1+C2+C3=1abcλ+2μμμλ+2μλμλ+2μμλ=λ+2μabc1μμ1λμ1μλR2R2-R1 , R3R3-R1=λ+2μabc1μμ0λ-μ000λ-μExpanding determinant along C1 is=1abc(λμ)2(λ+2μ)θ=2,β=2θ+β=4

 

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