F$=-($@DIV$\{$k@ABS$\{$q@Sub$\{0\}\}$@ABS$\{$q@Sub$\{3\}\}$;d@Sub$\{2\}$@Sup$\{2\}\})$@HAT$\{$j$\}-($@DIV$\{$k@ABS$\{$q@Sub$\{0\}\}$@ABS$\{$q@Sub$\{3\}\}$;d@Sub$\{2\}$@Sup$\{2\}\})$@HAT$\{$k$\}($Figure $3)$Now add a fourth charged particle, particle $3,$ with positive charge q$3,$ fixed in the yz$-$plane at $(0,$d$2,$d$2).$ What is the net force F$$ on particle $0$ due solely to this charge?

$($kabsq$0|$q$2|$d$22$kabsq$0|$q$1|$d$12)$j