Here,

$$\begin{aligned} \overrightarrow{\mathrm{AB}} & =(2-1) \hat{\mathbf{i}}+(-1-2) \hat{\mathbf{j}}+(4-3) \hat{\mathbf{k}} \\ & =\hat{\mathbf{i}}-3 \hat{\mathbf{j}}+\hat{\mathbf{k}} \end{aligned}$$

and

$$\begin{aligned} \overrightarrow{\mathbf{A C}} & =(4-1) \hat{\mathbf{i}}+(5-2) \hat{\mathbf{j}}+(-1-3) \hat{\mathbf{k}} \\ & =3 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-4 \hat{\mathbf{k}} \end{aligned}$$

$$\begin{aligned} & \therefore \quad \begin{aligned} \overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}} & =\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & -3 & 1 \\ 3 & 3 & -4 \end{array}\right| \\ & =\hat{\mathbf{i}}(12-3)-\hat{\mathbf{j}}(-4-3)+\hat{\mathbf{k}}(3+9) \\ & =9 \hat{\mathbf{i}}+7 \hat{\mathbf{j}}+12 \hat{\mathbf{k}} \end{aligned} \\ & \begin{aligned} \text { and } \quad |\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathbf{A C}}| & =\sqrt{9^2+7^2+12^2} \\ & =\sqrt{81+49+144} \\ & =\sqrt{274} \\ \therefore \quad \text { Area of } \triangle A B C & =\frac{1}{2}|\overrightarrow{\mathbf{A B}} \times \overrightarrow{\mathbf{A C}}| \\ & =\frac{1}{2} \sqrt{274} \text { sq units } \end{aligned} \end{aligned}$$

Let $A B C D$ and $A B F E$ are parallelograms on the same base $A B$ and between the same parallel lines $A B$ and $D F$.

Here, $A B \| C D$ and $A E \| B F$

Let $$\overrightarrow{A B}=\vec{a} \text { and } \overrightarrow{A D}=\vec{b}$$

$\therefore \quad$ Area of parallelogram $A B C D=\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}$

Now, area of parallelogram $A B F F=\overrightarrow{A B} \times \overrightarrow{A E}$

$$ \begin{aligned} &\begin{aligned} & =\overrightarrow{A B} \times(\overrightarrow{A D}+\overrightarrow{D E}) \\ & =\overrightarrow{A B} \times(\overrightarrow{\mathbf{b}}+k \overrightarrow{\mathbf{a}}) \quad[\text { let } \overrightarrow{D E}=k \overrightarrow{\mathbf{a}}, \text { where } k \text { is a scalar }] \\ & =\overrightarrow{\mathbf{a}} \times(\overrightarrow{\mathbf{b}}+k \overrightarrow{\mathbf{a}}) \\ & =(\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}})+(\overrightarrow{\mathbf{a}} \times k \overrightarrow{\mathbf{a}}) \\ & =(\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}})+k(\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{a}}) \\ & =(\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}) \quad[\because \overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{a}}=0] \\ & =\text { Area of parallelogram } A B C D \end{aligned}\\ &\text { Hence proved. } \end{aligned}$$

Here, components of C are c cos A and c sin A is drawn.

Since, $$\overrightarrow{C D}=b-c \cos A$$

In $\triangle B D C$,

$$a^2=(b-c \cos A)^2+(c \sin A)^2$$

$$\begin{array}{lrl} \Rightarrow & a^2 & =b^2+c^2 \cos ^2 A-2 b c \cos A+c^2 \sin ^2 A \\ \Rightarrow & 2 b c \cos A & =b^2-a^2+c^2\left(\cos ^2 A+\sin ^2 A\right) \\ \therefore & \cos A & =\frac{b^2+c^2-a^2}{2 b c} \end{array}$$

Since, $\overrightarrow{\mathbf{a}}, \overrightarrow{\mathbf{b}}$ and $\overrightarrow{\mathbf{c}}$ are the vertices of a $\triangle A B C$ as shown.

$$\begin{array}{ll} \therefore & \text { Area of } \triangle A B C=\frac{1}{2}|\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}| \\ \text { Now, } & \overrightarrow{\mathrm{AB}}=\overrightarrow{\mathbf{b}}-\overrightarrow{\mathbf{a}} \quad \text { and } \quad \overrightarrow{\mathrm{AC}}=\overrightarrow{\mathbf{c}}-\overrightarrow{\mathbf{a}} \\ \therefore & \text { Area of } \triangle A B C=\frac{1}{2}|\overrightarrow{\mathbf{b}}-\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{c}}-\overrightarrow{\mathbf{a}}| \end{array}$$

$$\begin{aligned} & =\frac{1}{2}|\vec{b} \times \vec{c}-\vec{b} \times \vec{a}-\vec{a} \times \vec{c}+\vec{a} \times \vec{a}| \\ & =\frac{1}{2}|\vec{b} \times \vec{c}+\vec{a} \times \vec{b}+\vec{c} \times \vec{a}+\overrightarrow{0}| \\ & =\frac{1}{2}|\vec{b} \times \vec{c}+\vec{a} \times \vec{b}+\vec{c} \times \vec{a}|\quad\text{... (i)} \end{aligned}$$

For three points to be collinear, area of the $\triangle A B C$ should be equal to zero.

$$\begin{array}{ll} \Rightarrow & \frac{1}{2}[\vec{b} \times \vec{c}+\vec{c} \times \vec{a}+\vec{a} \times \vec{b}]=0 \\ \Rightarrow & \vec{b} \times \vec{c}+\vec{c} \times \vec{a}+\vec{a} \times \vec{b}=0\quad\text{.... (ii)} \end{array}$$

This is the required condition for collinearity of three points $\overrightarrow{\mathbf{a}}, \overrightarrow{\mathrm{b}}$ and $\overrightarrow{\mathrm{c}}$.

$$\begin{aligned} &\text { Let } \hat{n} \text { be the unit vector normal to the plane of the } \triangle A B C \text {. }\\ &\begin{aligned} \therefore \quad \hat{n} & =\frac{\overrightarrow{A B} \times \overrightarrow{A C}}{|\overrightarrow{A B} \times \overrightarrow{A C}|} \\ & =\frac{\vec{a} \times \vec{b}+\vec{b} \times \vec{c}+\vec{c} \times \vec{a}}{|\vec{a} \times \vec{b}+\vec{b} \times \vec{c}+\vec{c} \times \vec{a}|} \end{aligned} \end{aligned}$$

Let ABCD be a parallelogram such that

$$\overrightarrow{A B}=\overrightarrow{\mathrm{p}}, \overrightarrow{\mathrm{AD}}=\overrightarrow{\mathrm{q}} \Rightarrow \overrightarrow{\mathrm{BC}}=\overrightarrow{\mathrm{q}}$$

By triangle law of addition, we get

$$\begin{aligned} & \overrightarrow{A C}=\vec{p}+\vec{q}=\vec{a} \quad\text{[say] ... (i)}\\ \text{Similarly,}\quad & \overrightarrow{B D}=-\vec{p}+\vec{q}=\vec{b}\quad\text{[say] ... (ii)} \end{aligned}$$

On adding Eqs. (i) and (ii), we get

$$\vec{a}+\vec{b}=2 \vec{q} \Rightarrow \vec{q}=\frac{1}{2}(\vec{a}+\vec{b})$$

On subtracting Eq. (ii) from Eq. (i), we get

$$\vec{a}-\vec{b}=2 \vec{p} \Rightarrow \vec{p}=\frac{1}{2}(\vec{a}-\vec{b})$$

$$\begin{aligned} \text{Now,}\quad \overrightarrow{\mathbf{p}} \times \overrightarrow{\mathbf{q}} & =\frac{1}{4}(\overrightarrow{\mathbf{a}}-\overrightarrow{\mathbf{b}}) \times(\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}) \\ & =\frac{1}{4}(\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}-\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{a}}-\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{b}}) \\ & =\frac{1}{4}[\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}] \\ & =\frac{1}{2}(\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}) \end{aligned}$$

So, area of a parallelogram $A B C D=|\overrightarrow{\mathbf{p}} \times \overrightarrow{\mathbf{q}}|=\frac{1}{2}|\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{b}}|$

$$\begin{aligned} &\text { Now, area of a parallelogram, whose diagonals are } 2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}} \text { and } \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-\hat{\mathbf{k}} \text {. }\\ &\begin{aligned} & =\frac{1}{2}|(2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}) \times(\hat{\mathbf{i}}+3 \hat{\mathbf{j}}-\hat{\mathbf{k}})| \\ & =\frac{1}{2}\left\|\begin{array}{rrr} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 2 & -1 & 1 \\ 1 & 3 & -1 \end{array}\right\| \\ & =\frac{1}{2}|[\hat{\mathbf{i}}(1-3)-\hat{\mathbf{j}}(-2-1)+\hat{\mathbf{k}}(6+1)]| \\ & =\frac{1}{2}|-2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+7 \hat{\mathbf{k}}| \\ & =\frac{1}{2} \sqrt{4+9+49} \\ & =\frac{1}{2} \sqrt{62} \text { sq units } \end{aligned} \end{aligned}$$