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

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

$$\begin{aligned} \text{Let}\quad & \overrightarrow{\mathbf{c}}=x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}} \\ & \text{Also,}\quad \overrightarrow{\mathbf{a}}=\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}} \text { and } \overrightarrow{\mathbf{b}}=\hat{\mathbf{j}}-\hat{\mathbf{k}} \end{aligned}$$

For $\overrightarrow{\mathbf{a}} \times \overrightarrow{\mathbf{c}}=\overrightarrow{\mathbf{b}}$,

$$\begin{aligned} & \Rightarrow \quad\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & 1 & 1 \\ x & y & z \end{array}\right|=\hat{\mathbf{j}}-\hat{\mathbf{k}} \\ & \Rightarrow \quad \hat{\mathbf{i}}(z-y)-\hat{\mathbf{j}}(z-x)+\hat{\mathbf{k}}(y-x)=\hat{\mathbf{j}}-\hat{\mathbf{k}} \end{aligned}$$

$$ \begin{aligned} \therefore \quad & z-y=0 \quad\text{.... (i)}\\ & x-z=1 \quad\text{.... (ii)}\\ & x-y=1\quad\text{.... (iii)} \end{aligned}$$

$$\begin{aligned} \text{Also,}\quad \overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{c}} & =3 \\ (\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}) \cdot(x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \mathbf{k}) & =3 \\ \Rightarrow\quad x+y+z & =3\quad \text{.... (iv)} \end{aligned}$$

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

$$2 x-y-z=2\quad\text{.... (v)}$$

$$\begin{aligned} &\text { On solving Eqs. (iv) and (v), we get }\\ &\begin{aligned} x & =\frac{5}{3} \\ \therefore\quad y & =\frac{5}{3}-1=\frac{2}{3} \text { and } z=\frac{2}{3} \\ \text{Now,}\quad \overrightarrow{\mathbf{c}} & =\frac{5}{3} \hat{\mathbf{i}}+\frac{2}{3} \hat{\mathbf{j}}+\frac{2}{3} \hat{\mathbf{k}} \\ & =\frac{1}{3}(5 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}) \end{aligned} \end{aligned}$$