Let the vectors given as

Solution:

We have,

$\vec{a}=a_{1} \hat{i}+a_{2} \hat{j}+a_{3} \hat{k}, \vec{b}=b_{1} \hat{i}+b_{2} \hat{j}+b_{3} \hat{k}, \vec{c}=c_{1} \hat{i}+c_{2} \hat{j}+c_{3} \hat{k}$

$(\vec{b}+\vec{c})=\left(b_{1}+c_{1}\right) \hat{i}+\left(b_{2}+c_{2}\right) \hat{j}+\left(b_{3}+c_{3}\right) \hat{k}$

Now, $\vec{a} \times(\vec{b}+\vec{c})\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ a_{1} & a_{2} & a_{3} \\ b_{1}+c_{1} & b_{2}+c_{2} & b_{3}+c_{3}\end{array}\right|$

$=\hat{i}\left[a_{2}\left(b_{3}+c_{3}\right)-a_{3}\left(b_{2}+c_{2}\right)\right]-\hat{j}\left[a_{1}\left(b_{3}+c_{3}\right)-a_{3}\left(b_{1}+c_{1}\right)\right]+\hat{k}\left[a_{1}\left(b_{2}+c_{2}\right)-a_{2}\left(b_{1}+c_{1}\right)\right]$

 

$=\hat{i}\left[a_{2} b_{3}+a_{2} c_{3}-a_{3} b_{2}-a_{3} c_{2}\right]+\hat{j}\left[-a_{1} b_{3}-a_{1} c_{3}+a_{3} b_{1}+a_{3} c_{1}\right]+\hat{k}\left[a_{1} b_{2}+a_{1} c_{2}-a_{2} b_{1}-a_{2} c_{1}\right]$

$\vec{a} \times \vec{b}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3}\end{array}\right|$     (2)

$=\hat{i}\left[a_{2} b_{3}-a_{3} b_{2}\right]+\hat{j}\left[b_{1} a_{3}-a_{1} b_{3}\right]+\hat{k}\left[a_{1} b_{2}-a_{2} b_{1}\right]$

$\vec{a} \times \vec{c}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ a_{1} & a_{2} & a_{3} \\ c_{1} & c_{2} & c_{3}\end{array}\right|$       (3)

On adding (2) and (3), we get:

$(\vec{a} \times \vec{b})+(\vec{a} \times \vec{c})=\hat{i}\left[a_{2} b_{3}+a_{2} c_{3}-a_{3} b_{2}-a_{3} c_{2}\right]+\hat{j}\left[b_{1} a_{3}+a_{3} c_{1}-a_{1} b_{3}-a_{1} c_{3}\right]$

$+\hat{k}\left[a_{1} b_{2}+a_{1} c_{2}-a_{2} b_{1}-a_{2} c_{1}\right]$         (4)

Now, from (1) and (4), we have:

$\vec{a} \times(\vec{b}+\vec{c})=\vec{a} \times \vec{b}+\vec{a} \times \vec{c}$

Hence, the given result is proved.