Answer
$x=0.366l$
$Q_2=0.402Q$
Work Step by Step
Assume the two charges are placed horizontally with up and right positive and left and down negative. Since both charges are negative, they will repulse and the third charge must be positive and in between.
$Q_1=-Q$
$Q_2=-3Q$
$Q_3$
$Q_1^-$_________$Q_2^+$_________$Q_3^-$
$\hspace{1.2cm}x\hspace{1.4cm}$$l-x$
$F_{21}=k\frac{Q_1Q_2}{r^2}=\big(9.0\times10^9\frac{Nm^2}{C^2}\big)\frac{(Q)Q_2}{x^2}$
$F_{23}=k\frac{Q_1Q_2}{r^2}=\big(9.0\times10^9\frac{Nm^2}{C^2}\big)\frac{Q_2(3Q)}{(l-x)^2}$
$F_{21}=F_{23}$
$\frac{QQ_2}{x^2}=\frac{Q_2(3Q)}{(l-x)^2}$
$\frac{1}{x^2}=\frac{3}{(l-x)^2}$
$x=\frac{l-x}{\sqrt{3}}$
$x=\frac{l}{\sqrt{3}+1}=0.366l$
$k\frac{QQ_2}{x^2}=k\frac{3Q^2}{l^2}$
$Q_2=\frac{3Qx^2}{l^2}=\frac{3}{\sqrt{3}+1}Q=0.402Q$
