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(C) For a system of linear equations to have no solution,the determinant of the coefficient matrix $\Delta$ must be $0$,and the system must be inconsi

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(C) For a system of linear equations to have no solution,the determinant of the coefficient matrix $\Delta$ must be $0$,and the system must be inconsistent.The coefficient matrix is $A = \begin{bmatrix} 2 & -3 & 5 \ 1 & 3 & -1 \ 3 & -1 & \lambda^2 - |\lambda| \end{bmatrix}$.Calculating the determinant $\Delta$:$\Delta = 2(3(\lambda^2 - |\lambda|) - 1) + 3(1(\lambda^2 - |\lambda|) - (-3)) + 5(-1 - 9)$$= 2(3\lambda^2 - 3|\lambda| - 1) + 3(\lambda^2 - |\lambda| + 3) + 5(-10)$$= 6\lambda^2 - 6|\lambda| - 2 + 3\lambda^2 - 3|\lambda| + 9 - 50$$= 9\lambda^2 - 9|\lambda| - 43$.Setting $\Delta = 0$ gives $9|\lambda|^2 - 9|\lambda| - 43 = 0$.Let $t = |\lambda|$. Then $9t^2 - 9t - 43 = 0$.The discriminant $D = (-9)^2 - 4(9)(-43) = 81 + 1548 = 1629 > 0$.Since $D > 0$,there are two distinct real roots for $t$. The roots are $t = \frac{9 \pm \sqrt{1629}}{18}$.Since $\sqrt{1629} \approx 40.36$,one root is positive and one is negative.Since $t = |\lambda| \ge 0$,only the positive root is valid.Thus,there is only $1$ value for $|\lambda|$,which corresponds to $2$ values of $\lambda$ (i.e.,$\lambda = \pm t$).

List-$I$	List-$II$
$(P)$ If $\beta=\frac{1}{2}(7\alpha-3)$ and $\gamma=28$,then the system has	$(1)$ a unique solution
$(Q)$ If $\beta=\frac{1}{2}(7\alpha-3)$ and $\gamma \neq 28$,then the system has	$(2)$ no solution
$(R)$ If $\beta \neq \frac{1}{2}(7\alpha-3)$ where $\alpha=1$ and $\gamma \neq 28$,then the system has	$(3)$ infinitely many solutions
$(S)$ If $\beta \neq \frac{1}{2}(7\alpha-3)$ where $\alpha=1$ and $\gamma=28$,then the system has	$(4)$ $x=11, y=-2$ and $z=0$ as a solution
	$(5)$ $x=-15, y=4$ and $z=0$ as a solution

The number of real values $\lambda$,such that the system of linear equations $2x - 3y + 5z = 9$,$x + 3y - z = -18$,and $3x - y + (\lambda^2 - |\lambda|)z = 16$ has no solution,is :-

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