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Question

If the system of equations $$ax+by+c=0$$ $$bx+cy+a=0$$  ,$$cx+ay+b=0$$ has a solution then the system of equations $$(b+c)x+(c+a)y+(a+b)z=0$$   ,$$(c+a)x+(a+b)y+(b+c)z=0$$  , $$(a+b)x+(b+c)y+(c+a)z=0$$ has 


A
only one solution
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B
no solution
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C
infinite number of solutions
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D
none of these
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Solution

The correct option is D infinite number of solutions
$$ax+by+c=0,bx+cy+a=0,cx+ay+b=0$$
Gives
$$\begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix}=0$$
Now $$(b+c)x+(c+a)y+(a+b)z=0,(c+a)x+(a+b)y+(b+c)z=0,(a+b)x+(b+c)y+(c+a)z=0$$
gives
$$\begin{vmatrix} b+c & c+a & a+b \\ c+a & a+b & b+c \\ a+b & b+c & c+a \end{vmatrix}=\begin{vmatrix} b & c & a \\ c & a & b \\ a & b & c \end{vmatrix}+\begin{vmatrix} c & a & b \\ a & b & c \\ b & c & a \end{vmatrix}=\begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix}+\begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix}=2\begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \end{vmatrix}=0$$

Mathematics

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