The system x+2y+3z=4,4x+5y+3z=5,3x+4y+3z=lamb | vedclass

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(B) The coefficient matrix $D$ is given by $\left|\begin{array}{lll}1 & 2 & 3 \ 4 & 5 & 3 \ 3 & 4 & 3\end{array}ight|$.Calculating the determinant

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-$86$

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(B) The coefficient matrix $D$ is given by $\left|\begin{array}{lll}1 & 2 & 3 \ 4 & 5 & 3 \ 3 & 4 & 3\end{array}ight|$.Calculating the determinant: $1(15-12) - 2(12-9) + 3(16-15) = 1(3) - 2(3) + 3(1) = 3 - 6 + 3 = 0$.Since $D=0$,for the system to be consistent,the Cramer's rule determinants $D_1, D_2, D_3$ must also be $0$.Calculating $D_3 = \left|\begin{array}{lll}1 & 2 & 4 \ 4 & 5 & 5 \ 3 & 4 & \lambda\end{array}ight| = 0$.$1(5\lambda - 20) - 2(4\lambda - 15) + 4(16 - 15) = 0$.$5\lambda - 20 - 8\lambda + 30 + 4 = 0$.$-3\lambda + 14 = 0 \Rightarrow 3\lambda = 14$.Given $3\lambda = n + 100$,we substitute $3\lambda = 14$:$14 = n + 100 \Rightarrow n = 14 - 100 = -86$.

List-$I$	List-$II$
$(A)$ $\lambda=8, \mu \neq 15$	$1$. Infinitely many solutions
$(B)$ $\lambda \neq 8, \mu \in R$	$2$. No solution
$(C)$ $\lambda=8, \mu=15$	$3$. Unique solution

The system $x+2y+3z=4$,$4x+5y+3z=5$,$3x+4y+3z=\lambda$ is consistent and $3\lambda=n+100$,then $n=$

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