Consider the given equation of lines,
$x-(t+\alpha)=0$ ...(i)
$y+16=0$ ...(ii)
and $-\alpha x +y=0$ ...(iii)
Since, these lines are concurrent, therefore the system of equations is consistent.
Now, $\begin{vmatrix} 1 & 0 & -(t+\alpha) \\ 0 & 1 & 16 \\ -\alpha & 1 & 0 \end{vmatrix}=0$
$\Rightarrow 1(0-16)-(t+\alpha)(0+\alpha)=0$
$\Rightarrow -16-\alpha(t+\alpha)=0$
$\Rightarrow \alpha(t+\alpha)+16=0$
$\Rightarrow \alpha^{2}+t \alpha+16=0$
Clearly, $\alpha$ should be real.
$\therefore t^{2}-4 \times 16 \geq 0$
$\Rightarrow t^{2}-64 \geq 0$
Hence, least positive value of $t$ is 8.