Solve this

Using the equations, we get

$D=\left|\begin{array}{ccc}1 & -1 & 3 \\ 1 & 3 & -3 \\ 5 & 3 & 3\end{array}\right|=1(9+9)+1(3+15)+3(3-15)$

$=18+18-36=0$

$D_{1}=\left|\begin{array}{ccc}6 & -1 & 3 \\ -4 & 3 & -3 \\ 10 & 3 & 3\end{array}\right|=6(9+9)+1(-12+30)+3(-12-30)$

$=108+18-126=0$

$D_{2}=\left|\begin{array}{ccc}1 & 6 & 3 \\ 1 & -4 & -3 \\ 5 & 10 & 3\end{array}\right|=1(-12+30)-6(3+15)+3(10+20)$

$=18-108+90=0$

$D_{3}=\left|\begin{array}{ccc}1 & -1 & 6 \\ 1 & 3 & -4 \\ 5 & 3 & 10\end{array}\right|=1(30+12)+1(10+20)+6(3-15)$

$=42+30-72=0$

$\therefore D=D_{1}=D_{2}=D_{3}=0$

Hence, the system of equations has infinitely many solutions.