Given system of linear equations is
\(\begin{aligned}
x+2 y+3 z & =6 \\
x+3 y+5 z & =9 \\
\text { and } \quad 2 x+5 y+\lambda z & =\mu
\end{aligned}\)
Now, according to Cramer's rule,
\(\begin{aligned}
& \Delta=\left|\begin{array}{lll}
1 & 2 & 3 \\
1 & 3 & 5 \\
2 & 5 & \lambda
\end{array}\right|=1(3 \lambda-25)-2(\lambda-10)+3(5-6) \\
& =\lambda-8 \\
& \Delta_1=\left|\begin{array}{lll}
6 & 2 & 3 \\
9 & 3 & 5 \\
\mu & 5 & \lambda
\end{array}\right|=6(3 \lambda-25)-2(9 \lambda-5 \mu) \\
& +3(45-3 \mu)=\mu-15 \\
& \Delta_2=\left|\begin{array}{lll}
1 & 6 & 3 \\
1 & 9 & 5 \\
2 & \mu & \lambda
\end{array}\right|=1(9 \lambda-5 \mu)-6(\lambda-10)+3(\mu-18) \\
& =3 \lambda-4 \mu+6
\end{aligned}\)
and
\(\Delta_3=\left|\begin{array}{lll}
1 & 2 & 6 \\
1 & 3 & 9 \\
2 & 5 & \mu
\end{array}\right|=1(3 \mu-45)-2(\mu-18)+6(5-6) = \mu - 15\)
Now, if \(\lambda=8\) and \(\mu \neq 15\), then system of linear equations has no solution.
If \(\lambda \neq 8\) and \(\mu \in R\), then system of linear equations has unique solution.
And, if \(\lambda=8\) and \(\mu=15\), then system of linear equations has infinite number of solutions, because \(\Delta_2=3 \lambda-2 \mu+6\) is also be zero. Hence, option (2) is correct.