Answer
$\tau = (6.0~N\cdot m)~\hat{i}-(3.0~N\cdot m)~\hat{j}-(6.0~N\cdot m)~\hat{k}$
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 jar's position in unit-vector notation:
$r = (3.0~m)~\hat{i}- (2.0~m)~\hat{j}+ (4.0~m)~\hat{k}$
We can express the force in unit-vector notation:
$F =(3.0~N)~\hat{i}-(4.0~N)~\hat{j}+(5.0~N)~\hat{k}$
We can find the torque about the origin on the jar:
$\tau = r \times F$
$\tau = [(-2.0~m)(5.0~N)-(-4.0~N)(4.0~m)]~\hat{i}+[(4.0~m)(3.0~N)-(5.0~N)(3.0~m)]~\hat{j}+[(3.0~m)(-4.0~N)-(3.0~N)(-2.0~m)]~\hat{k}$
$\tau = (6.0~N\cdot m)~\hat{i}-(3.0~N\cdot m)~\hat{j}-(6.0~N\cdot m)~\hat{k}$
