$$\begin{gathered} \mathbf{Three}\mathbf{}\mathbf{vectors}\mathbf{}\mathbf{are}\mathbf{}\mathbf{given}\mathbf{}\mathbf{by}\mathbf{}\mathbf{a}\mathbf{}\mathbf{⃗}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{0}\hat{\mathbf{i}}\mathbf{+}\mathbf{3}\mathbf{.}\mathbf{0}\hat{\mathbf{j}}\mathbf{-}\mathbf{2}\mathbf{.}\mathbf{0}\hat{\mathbf{k}}\overrightarrow{\mathbf{a}}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{0}\hat{\mathbf{i}}\mathbf{+}\mathbf{3}\mathbf{.}\mathbf{0}\hat{\mathbf{j}}\mathbf{-}\mathbf{2}\mathbf{.}\mathbf{0}\hat{\mathbf{k}}\mathbf{}\overrightarrow{\mathbf{a}} \\ \mathbf{=}\mathbf{3}\mathbf{.}\mathbf{0}\hat{\mathbf{i}}\mathbf{+}\mathbf{3}\mathbf{.}\mathbf{0}\hat{\mathbf{j}}\mathbf{-}\mathbf{2}\mathbf{.}\mathbf{0}\hat{\mathbf{k}}\mathbf{,}\overrightarrow{\mathbf{b}}\mathbf{=}\mathbf{-}\mathbf{1}\mathbf{.}\mathbf{0}\hat{\mathbf{i}}\mathbf{-}\mathbf{4}\mathbf{.}\mathbf{0}\hat{\mathbf{j}}\mathbf{+}\mathbf{2}\mathbf{.}\mathbf{0}\hat{\mathbf{k}}\mathbf{,}\mathbf{}\overrightarrow{\mathbf{c}}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{0}\hat{\mathbf{i}}\mathbf{+}\mathbf{2}\mathbf{.}\mathbf{0}\hat{\mathbf{j}}\mathbf{+}\mathbf{1}\mathbf{.}\mathbf{0}\widehat{\mathbf{k}\mathbf{.}} \\ \mathbf{Find}\mathbf{}\mathbf{\left(}\mathbf{a}\mathbf{\right)}\overrightarrow{\mathbf{a}}\mathbf{.}\mathbf{(}\overrightarrow{\mathbf{a}}\mathbf{\times }\overrightarrow{\mathbf{c}}\mathbf{)}\mathbf{,}\mathbf{\left(}\mathbf{b}\mathbf{\right)}\mathbf{}\overrightarrow{\mathbf{a}}\mathbf{.}\mathbf{(}\overrightarrow{\mathbf{b}}\mathbf{+}\overrightarrow{\mathbf{c}}\mathbf{)}\mathbf{}\mathbf{,}\mathbf{\left(}\mathbf{c}\mathbf{\right)}\mathbf{}\overrightarrow{\mathbf{a}}\mathbf{\times }\mathbf{(}\overrightarrow{\mathbf{b}}\mathbf{+}\overrightarrow{\mathbf{c}}\mathbf{)}\mathbf{.} \end{gathered}$$

Vector operations can be used to find the dot product, cross product, and addition between two vectors. The addition of vectors gives another vector quantity. The cross product of two vectors results in a vector quantity that is perpendicular to both vectors whereas the dot product of two vectors produces a scalar quantity.

The formula for the cross product is,

$\overrightarrow{a}\times \overrightarrow{b}=(a_y{b}_z-b_y{a}_z)\hat{i}+(a_z{b}_x-b_z{a}_x)\hat{j}+(a_x{b}_y-b_x{a}_y)\hat{k}$ (1)

The formula for the dot product is,

$\overrightarrow{a}.\overrightarrow{b}=a\times b\times \cos \theta$ (2)

First, find the individual values of each term like $(\overrightarrow{\mathrm{b}}\times \overrightarrow{\mathrm{c}})$and $(\overrightarrow{b}+\overrightarrow{c})$. After that using the formula for the dot and cross product of vectors, get the required answer.

The vectors are given below:

$$\begin{gathered} \overrightarrow{a}=3.0\hat{i}+3.0\hat{j}-2.0\hat{k} \\ \hat{b}=-1.0\hat{i}-4.0\hat{j}+2.0\hat{k} \\ \overrightarrow{c}=2.0\hat{i}+2.0\hat{j}+1.0\hat{k} \end{gathered}$$

$$\begin{gathered} \mathrm{Write}\mathrm{the}\mathrm{equation}\left(\mathrm{i}\right)\mathrm{for}\overrightarrow{\mathrm{b}}\mathrm{and}\overrightarrow{\mathrm{c}}. \\ (\overrightarrow{\mathrm{b}}\times \overrightarrow{\mathrm{c}})=({\mathrm{b}}_{\mathrm{y}}{\mathrm{c}}_{\mathrm{z}}-{\mathrm{c}}_{\mathrm{y}}{\mathrm{b}}_{\mathrm{z}})\hat{\mathrm{i}}+({\mathrm{b}}_{\mathrm{z}}{\mathrm{c}}_{\mathrm{x}}-{\mathrm{b}}_{\mathrm{x}}{\mathrm{c}}_{\mathrm{z}})\hat{\mathrm{j}}+({\mathrm{b}}_{\mathrm{x}}{\mathrm{c}}_{\mathrm{y}}-{\mathrm{b}}_{\mathrm{y}}{\mathrm{c}}_{\mathrm{x}})\hat{\mathrm{k}} \\ =-8.0\hat{\mathrm{i}}+5.0\hat{\mathrm{j}}+6.0\hat{\mathrm{k}} \\ \mathrm{Find}\mathrm{the}\mathrm{dot}\mathrm{product}\mathrm{of}\overrightarrow{\mathrm{b}}\times \overrightarrow{\mathrm{c}}\mathrm{with}\overrightarrow{\mathrm{a}}. \\ \overrightarrow{\mathrm{a}}.(\overrightarrow{\mathrm{b}}\times \overrightarrow{\mathrm{c}})=\left(\right(3.0\times 8.0\left)\right)+\left(\right(3.0\times 5.0\left)\right)+\left(\right(-2.0\times 6.0\left)\right) \\ =-21 \\ \mathrm{Thus},\mathrm{the}\mathrm{value}\mathrm{of}\overrightarrow{\mathrm{a}}.(\overrightarrow{\mathrm{b}}\times \overrightarrow{\mathrm{c}})\mathrm{is}-21. \end{gathered}$$

$$\begin{gathered} \mathrm{First},\mathrm{find}\mathrm{the}\mathrm{value}\mathrm{of}\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}. \\ (\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}})=(-1.0+2.0)\hat{\mathrm{i}}+(-4.0+2.0)\hat{\mathrm{j}}+(2.0+1.0)\hat{\mathrm{k}} \\ =1.0\hat{\mathrm{i}}+2.0\hat{\mathrm{j}}+3.0\hat{\mathrm{k}} \\ (\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}})=1.0\hat{\mathrm{i}}+2.0\hat{\mathrm{j}}+3.0\hat{\mathrm{k}}\left(3\right) \\ \mathrm{Now},\mathrm{find}\mathrm{the}\mathrm{dot}\mathrm{product}\mathrm{of}\overrightarrow{\mathrm{a}}\mathrm{with}\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}. \\ \overrightarrow{\mathrm{a}}.(\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}})=\left(\right(3.0\times 1.0\left)\right)+\left(\right(3.0\times -2.0\left)\right)+\left(\right(-2.0\times 3.0\left)\right) \\ =-9.0 \\ \mathrm{Thus},\mathrm{the}\mathrm{value}\mathrm{of}\overrightarrow{\mathrm{a}}.(\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}})\mathrm{is}-9.0. \end{gathered}$$

$$\begin{gathered} \mathrm{Use}\mathrm{the}\mathrm{equation}\left(\mathrm{iii}\right)\mathrm{to}\mathrm{write}\mathrm{the}\mathrm{cross}\mathrm{product}\mathrm{of}\overrightarrow{\mathrm{a}}\mathrm{and}\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}. \\ \overrightarrow{\mathrm{a}}\times (\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}})=\left(3.0\hat{\mathrm{i}}+3.0\hat{\mathrm{j}}-2.0\hat{\mathrm{k}}\right)\times \left(1.0\hat{\mathrm{i}}-2.0\hat{\mathrm{j}}+3.0\hat{\mathrm{k}}\right) \\ \mathrm{Calculate}\mathrm{the}\mathrm{cross}\mathrm{product}\mathrm{using}\mathrm{equation}\left(\mathrm{i}\right) \\ \overrightarrow{\mathrm{a}}\times (\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}})=5.0\hat{\mathrm{i}}-11\hat{\mathrm{j}}-9.0\hat{\mathrm{k}} \\ \mathrm{Therefore},\mathrm{the}\mathrm{value}\mathrm{of}\overrightarrow{\mathrm{a}}\times (\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}})\mathrm{is}5\hat{\mathrm{i}}-11\hat{\mathrm{j}}-9\hat{\mathrm{k}} \end{gathered}$$