If the system of equations $ax+ by + c= 0, bx+ xy + a = 0 , cx + ay + b = 0$ has infinitely many solutions then the system of equations

$(b+c) x+(c+a) y +(a+z) z=0$

$(c+a)x+(a+b)y+(b+c) z=0$

$(a+b)x+(b+c) y+(c+a)z=0$ has

infinite number of solutions

The correct answer is option (3) : infinite number of solutions

The system of equations

$ax+ by + c= 0 $

$bx+ cy + a= 0 $

$cx+ ay + b = 0 $ has inginitely many solutions

$∴\begin{bmatrix}a & b & c\\b & c & a\\c & a & b\end{bmatrix}=0$ ..........................(i)

Now,

$\begin{bmatrix}b+c & c+a & a+b\\c+a & a+b & b+c\\a+b & b+c & c+a\end{bmatrix}$

$=\begin{bmatrix}2(a+b+c) & c+a & a+b\\2(a+b+c)& a+b & b+c\\2(a+b+c) & b+c & c+a\end{bmatrix}$ [Applying $C_1→C_1+C_2+C_3$]

$=2\begin{bmatrix}a+b+c & c+a & a+b\\a+b+c& a+b & b+c\\a+b+c & b+c & c+a\end{bmatrix}$

$=2\begin{bmatrix}a+b+c & -b & -c\\a+b+c& -c & -a\\a+b+c & -a & -b\end{bmatrix}$           $\begin{bmatrix} Applying \, \, C_2→C_2-C_1\\C_3→C_3-C_1\end{bmatrix}$

$=2\begin{bmatrix}a & -b & -c\\b & -c & -a\\c & -a & -b\end{bmatrix}$     [Applying $C_1→C_1+C_2+C_3$]

$=2\begin{bmatrix}a & b & c\\b & c & a\\c & a & b\end{bmatrix}= 2× 0 = 0 $         [Using (i) ]

Hence, the system of equations has infinitely many solutions.