Answer
$F_x= -5.00~N$
Work Step by Step
We can write the general form of a cross-product:
$a\times b = (a_yb_z-b_ya_z)~\hat{i}+(a_zb_x-b_za_x)~\hat{j}+(a_xb_y-b_xa_y)~\hat{k}$
We can express the particle's position in unit-vector notation:
$r = (2.00~m)~\hat{i}+(-3.00~m)~\hat{j}+ (2.00~m)~\hat{k}$
We can express the force in unit-vector notation:
$F = F_x~\hat{i}+(7.00~N)~\hat{j}+(-6.00~N)~\hat{k}$
We can find the torque about the origin on the particle:
$\tau = r \times F$
$\tau = [(-3.00~m)(-6.00~N)-(7.00~N)(2.00~m)]~\hat{i}+ [(2.00~m)F_x-(-6.00~N)(2.00~m)]~\hat{j}+ [(2.00~m)(7.00~N)-F_x~(-3.00~m)]~\hat{k}$
$\tau = (4.00~N\cdot m)~\hat{i}+ (2.00~F_x+12.0~N\cdot m)~\hat{j}+ (14.0~N\cdot m+3.00~F_x)~\hat{k}$
It is given that:
$\tau = (4.00~N\cdot m)~\hat{i}+ (2.00~N\cdot m)~\hat{j}-(1.00~N\cdot m)~\hat{k}$
Therefore:
$2.00~F_x+12.0~N\cdot m = 2.00~N\cdot m$
$2.00~F_x= -10.0~N\cdot m$
$F_x= -5.00~N$
