Mathematics - Determinant Question with Solution | TestHub

Given, the system of equations

$x+y+z=6$

$x+2y+\alpha z=10$

$x+3y+5z=\beta$

Now finding,

$△=\left|\begin{array}{ccc}1& 1& 1\\ 1& 2& \alpha \\ 1& 3& 5\end{array}\right|$

$\Rightarrow △=1\left(10-3\alpha \right)-\left(5-\alpha \right)+\left(3-2\right)$

$\Rightarrow △=6-2\alpha$

Now for unique solution, $△\ne 0\Rightarrow \alpha \ne 3$

So, for unique solution $\Rightarrow \alpha \ne 3$

But in option $\left(D\right)$, system has unique solution for $\alpha =3$ which is completely wrong.