Answer
$\tau = (-8.00~N\cdot m)~\hat{i}-(26.0~N\cdot m)~\hat{j}-(40.0~N\cdot m)~\hat{k}$
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 object's position in unit-vector notation:
$d = (2.00~m)~\hat{i}+ (4.00~m)~\hat{j}- (3.00~m)~\hat{k}$
We can express the force in unit-vector notation:
$F = (6.00~N)~\hat{i}-(8.00~N)~\hat{j}+(4.00~N)~\hat{k}$
We can find the torque about the origin:
$\tau = d \times F$
$\tau = [(4.00~m)(4.00~N)-(-8.00~N)(-3.00~m)]~\hat{i}+[(-3.00~m)(6.00~N)-(4.00~N)(2.00~m)]~\hat{j}+[(2.00~m)(-8.00~N)-(6.00~N)(4.00~m)]~\hat{k}$
$\tau = (-8.00~N\cdot m)~\hat{i}-(26.0~N\cdot m)~\hat{j}-(40.0~N\cdot m)~\hat{k}$
