If the system of equations \(x+5y+6z=4\) \(2x+3y+4z=7\), \(x+6y+az=b\) has infinitely many solutions, then the point \((…

\( \Delta=\left|\begin{array}{lll}
1 & 5 & 6 \\
2 & 3 & 4 \\
1 & 6 & a
\end{array}\right|=0 \)
\( \Rightarrow(3 a-24)-5(2 a-4)+6(9)=0 \)
\( \Rightarrow-7 a+50=0 \)
\( a=\frac{50}{7}\)
Now \( \Delta_z=0 \Rightarrow\left|\begin{array}{lll}
1 & 5 & 4 \\
2 & 3 & 7 \\
1 & 6 & b
\end{array}\right|=0 \)
\(\Rightarrow(3 b-42)-5(2 b-7) +4(9)=0 \)
\( \Rightarrow-7 b+29=0\)
\(b=\frac{29}{7} \)
\( a-b=3 \)
\( \therefore x-y=3\)