Answer
a) $q_1$ adds a field that points to the left
b) $x=0.297m$
Work Step by Step
(a) We know that if only $q_2$, $q_3$ were present then the electric field would be zero at $x=0.30m$. Charge $q_1$ adds a field that points to the left for all positive x and it cancels the combined fields from $q_2$ and $q_3$.
(b) We know that
$\vec{E}=-\frac{Kq}{x^2}\hat x-\frac{Kq}{(0.40-x)^2}\hat x+\frac{Kq}{(x-0.20)^2}\hat x=0$
$\implies \frac{E}{Kq}=[-\frac{1}{x^2}-\frac{1}{(0.40-x)^2}+\frac{1}{(x-0.20)^2}]=0$
This simplifies to:
$x=0.297m$
