For the system of equations x + y + z = 6 x + 2y + = 10 x + 3y + 5z...

Given, the system of equations

$x + y + z = 6$

$x + 2y + \alphaz = 10$

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

Now finding,

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

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

$⇒ △ = 6 - 2\alpha$

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

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

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