Correct option is : (1) 25
\(3 x+5 y+\lambda z=3\)
\(7 x+11 y-9 z=2\)
\(97 x+155 y-189 z=\mu \)
\(\text { By cramer's rule, } \Delta=\Delta_1=\Delta_2=\Delta_3=0\)
\( \left|\begin{array}{ccc}
3 & 5 & \lambda \\
7 & 11 & -9 \\
97 & 155 & -189
\end{array}\right|=0\)
\( \mathrm{R}_3 \rightarrow \mathrm{R}_3-14 \mathrm{R}_2\)
\(\left|\begin{array}{ccc}
3 & 5 & \lambda \\
7 & 11 & -9 \\
-1 & 1 & -63
\end{array}\right|=0\)
\(\mathrm{C}_1 \rightarrow \mathrm{C}_1+\mathrm{C}_2\)
\( 2\left|\begin{array}{ccc}
4 & 5 & \lambda \\
9 & 11 & -9 \\
0 & 1 & -63
\end{array}\right|=0 \)
\( -1(-36-9 \lambda)-63(44-45)=0\)
\(36+9 \lambda+63=0\)
\( 9 \lambda=-99 \)
\(\lambda=-11\)
\(\Delta_3=0\)
\( \left|\begin{array}{ccc}
3 & 5 & 3 \\
7 & 11 & 2 \\
97 & 155 & \mu
\end{array}\right|=0 \)
\( \mathrm{C}_2 \rightarrow \mathrm{C}_2-\mathrm{C}_1\)
\( \left|\begin{array}{ccc}
3 & 2 & 3 \\
7 & 4 & 2 \\
97 & 58 & \mu
\end{array}\right|=0\)
\(C_1 \rightarrow C_1-C_3\)
\( \left|\begin{array}{ccc}
0 & 2 & 3 \\
5 & 4 & 2 \\
97-\mu & 58 & \mu
\end{array}\right|=0\)
\( -5(2 \mu-174)+(97-\mu)(4-12)=0 \)
\(10 \mu-870+776-8 \ \mu=0 \)
\(2 \mu=94 \)
\( \mu=47\)
\(\mu+2 \lambda=47-22\)
= 25
