#### Explain solution RD Sharma class 12 chapter Determinants exercise multiple choise question 12 maths

Correct option (b)

Hint:

First solve the given determinant, then conclude answer.

Given:

$\left | a\, b\, 2a\alpha +3b\: b\: c\: 2b\alpha +3c\: 2a\alpha +3b\: 2b\alpha +3c\: 0 \right |=0$

We have to find the conclusion of determinant

Solution:

$\begin{vmatrix} a &b &2a\alpha +3b \\ b &c &2b\alpha +3c \\ 2a\alpha +3b &2b\alpha +3c &0 \end{vmatrix}=0$

Expanding along R1, we have

$a\left[-(2 b \alpha+3 c)^{2}\right]-b[-(2 b \alpha+3 c)(2 a \alpha+3 b)]+(2 a \alpha+3 b)[b(2 b \alpha+3 c)-c(2 a \alpha+3 b)]=0$

$\Rightarrow -a(2b\alpha +3c)^{2}+b(2b\alpha +3c)(2a\alpha +3b)+(2a\alpha +3b)(2b^{2}\alpha +3bc-2ac\alpha -3bc)=0$

$\Rightarrow -a(2b\alpha +3c)^{2}+b(2b\alpha +3c)(2a\alpha +3b)+(2a\alpha +3b)(2\alpha (b^{2}-ac))=0$

$\Rightarrow (2b\alpha +3c)[-2ab\alpha -3ac+2ab\alpha +3b^{2}]+(2a\alpha +3b)(2\alpha )(b^{2}-ac)=0$

$\Rightarrow (2b\alpha +3c)[3b^{2}-3ac]+(2a\alpha +3b)(2\alpha )(b^{2}-ac)=0$

$\Rightarrow (b^{2}-ac)[6b\alpha +9c+4a\alpha ^{2}+6b\alpha ]=0$

$\Rightarrow (b^{2}-ac)[4a\alpha ^{2}+12b\alpha +9c ]=0$

$\Rightarrow (b^{2}-ac)=0\; \; or\; \;[4a\alpha ^{2}+12b\alpha +9c ]=0$

$\Rightarrow a,b,c\: are in\; G.P\; or\; [4a\alpha ^{2}+12b\alpha +9c ]=0\; \; whose\; root\; is\; \alpha .$

$Hence\; \alpha \; is\;a\; root\; o\! f\; [4a\alpha ^{2}+12b\alpha +9c ]=0\; \; or\; a,b,c\; are\; in\; G.P .$