Brightest bulb

The brightness of bulb $A$ is proportional to $P$ the power consumed by it or $I^2$ the square of the current flowing through it. Let the supply voltage be $V$ and the resistance of the bulb be $R$ , then the current $I$ is given by:

$I = \cfrac{V}{R+\cfrac{R}{n}}$

where $n = 0, 1, 2, 3$ is the number of switches $S_1, S_2, S_3$ are closed.

$I$ is maximum when $\left(R+\cfrac{R}{n}\right)$ is the lowest, that is when $n=3$ .

Therefore, bulb $A$ is the brightest when $\boxed{all\space S_1, S_2,\space S_3\space are\space all\space closed}$ .