Step 1

We are given the force $F = \left\{ {2\hat i + 4\hat j - 6\hat k} \right\}\;{\rm{N}}$.

We are asked to express the moment of the force about point P in Cartesian vector form.

 
Step 2

The coordinates of point O, A,and P can be written as:

\begin{array}{c} O\left( {0,\;0,\;0} \right)\;{\rm{m}}\\ A\left( {3,\;3,\; - 1} \right)\;{\rm{m}}\\ P\left( { - 2,\; - 3,\;2} \right)\;{\rm{m}} \end{array}

The position vector of point O can be written as:

\[{r_O} = \left( {0\hat i + 0\hat j + 0\hat k} \right)\;{\rm{m}}\]

The position vector of point A can be written as:

\[{r_A} = \left( {3\hat i + 3\hat j - 1\hat k} \right)\;{\rm{m}}\]

Similarly, the position vector of point P can be written as:

\[{r_P} = \left( { - 2\hat i - 3\hat j + 2\hat k} \right)\;{\rm{m}}\]
 
Step 3

To calculate the position vector from point O to point A,we have:

\begin{array}{c} {r_{OA}} = {r_A} - {r_O}\\ {r_{OA}} = \left( {3\hat i + 3\hat j - 1\hat k} \right)\;{\rm{m}} - \left( {0\hat i + 0\hat j + 0\hat k} \right)\;{\rm{m}}\\ {r_{OA}} = \left( {3\hat i + 3\hat j - 1\hat k} \right)\;{\rm{m}} \end{array}

To calculate the position vector from point O to point P, we have:

\begin{array}{c} {r_{OP}} = {r_P} - {r_O}\\ {r_{OP}} = \left( { - 2\hat i - 3\hat j + 2\hat k} \right)\;{\rm{m}} - \left( {0\hat i + 0\hat j + 0\hat k} \right)\;{\rm{m}}\\ {r_{OP}} = \left( { - 2\hat i - 3\hat j + 2\hat k} \right)\;{\rm{m}} \end{array}

To calculate the position vector from point P to point A,we have:

\begin{array}{c} {r_{PA}} = {r_{OA}} - {r_{OP}}\\ {r_{PA}} = \left( {3\hat i + 3\hat j - 1\hat k} \right)\;{\rm{m}} - \left( { - 2\hat i - 3\hat j + 2\hat k} \right)\;{\rm{m}}\\ {r_{PA}} = \left( {5\hat i + 6\hat j - 3\hat k} \right)\;{\rm{m}} \end{array}
 
Step 4

To calculate the moment of force $F$ about point P, we have:

\[{M_P} = {r_{PA}} \times F\]

Substitute the values in the above expression, we get:

\[\begin{array}{c} {M_P} = \left( {5\hat i + 6\hat j - 3\hat k} \right)\;{\rm{m}} \times \left( {2\hat i + 4\hat j - 6\hat k} \right)\;{\rm{kN}}\\ {M_P} = \left| {\begin{array}{*{20}{c}} {\hat i}&{\hat j}&{\hat k}\\ 5&6&{ - 3}\\ 2&4&{ - 6} \end{array}} \right|\;{\rm{kN}} \cdot {\rm{m}}\\ {M_P} = \left[ \begin{array}{l} \hat i\left( {\left( { - 6 \times 6} \right) + \left( {4 \times 3} \right)} \right) - \hat j\left( {\left( { - 6 \times 5} \right) + \left( {3 \times 2} \right)} \right)\\ + \hat k\left( {\left( {5 \times 4} \right) - \left( {2 \times 6} \right)} \right) \end{array} \right]\;{\rm{kN}} \cdot {\rm{m}}\\ {M_P} = \left( { - 24\hat i + 24\hat j + 8\hat k} \right)\;{\rm{kN}} \cdot {\rm{m}} \end{array}\]