Question:
The value of $\mathrm{k} \in \mathbf{R}$, for which the following system of linear equations
$3 x-y+4 z=3$
$x+2 y-3 z=-2$
$6 x+5 y+k z=-3$
has infinitely many solutions, is :
Correct Option: , 2
Solution:
$\left|\begin{array}{ccc}3 & -1 & 4 \\ 1 & 2 & -3 \\ 6 & 5 & \mathrm{~K}\end{array}\right|=0$
$\Rightarrow 3(2 \mathrm{~K}+15)+\mathrm{K}+18-28=0$
$\Rightarrow 7 \mathrm{~K}+35=0 \Rightarrow \mathrm{K}=-5$
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