If the system of equations $ax+by+c=0, bx + cy+a=0, cx+ay+b=0$ has infinitely many solutions 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
Answer & explanation
Correct answer: option 3
The system of equations
$ax + by + c = 0$
$bx + cy + a=0$
$cx+ay+b=0$ has infinitely many solutions
$∴\begin{vmatrix}a &b& c\\b&c &a\\c&a&b\end{vmatrix}=0$ ...(i)
Now,
$\begin{vmatrix}b+c &c+a &a+b\\c+a &a+b &b+c\\a+b &b+c &c+a\end{vmatrix}$
$=\begin{vmatrix}2(a+b+c) &c+a &a+b\\2(a+b+c) &a+b &b+c\\2(a+b+c) &b+c &c+a\end{vmatrix}$ [Applying $C_1→C_1+C_2 + C_3$]
$=2\begin{vmatrix}a+b+c &c+a &a+b\\a+b+c &a+b &b+c\\a+b+c &b+c &c+a\end{vmatrix}$
$=2\begin{vmatrix}a+b+c &-b &-c\\a+b+c &-c &-a\\a+b+c &-a &-b\end{vmatrix}$ [Applying $C_2 →C_2-C_1,C_3→ C_3-C_1$]
$=2\begin{vmatrix}a &-b &-c\\b &-c &-a\\c &-a &-b\end{vmatrix}$ [Applying $C_1→C_1 + C_2 + C_3$]
$=2\begin{vmatrix}a &b &c\\b &c &a\\c &a &b\end{vmatrix}=2×0$ [Using (i)]
Hence, the system of equations has infinitely many solutions.