Given :- $\vec F = 80i+50j-60k \\ \vec{r_A } = 6i+2j+6k \\ \vec {r_B } = 8i+j+4k $

To find:-

Moment about $B \vec{ (M_B ) }$

Solution:-

Position vector of A w.r.t B is

$\vec{r_{AB} }= \vec{r_A } - \vec{r_B }\\ =(6i+2j+6k)-(8i+j+4k) \\ \vec{r_{AB}} = -2i+j+2k \\ Now, \\ \vec{M_B } = \vec{r_{AB}} × \vec F$

$$= \begin{vmatrix}i& j& k\\ -2 & 1&2\\ 80&50 &-60 \end{vmatrix}$$

$ =(-60-100)i-(120-160)j+(-100-80)k \Rightarrow Ans \\ \therefore \vec{M_B} =-160i+40j-180k \\ |\vec{M_B} |=\sqrt{(-160)^2+(40)^2+(-180)^2}\\ \therefore |\vec {M_B} |=244.13 Units ---- Ans$