The area of the parallelogram whose adjacent sides are $\hat{i} + \hat{k}$ and $2\hat{i}+\hat{j}+\hat{k}$ is

$\sqrt{3}$

$\vec{a} = \hat{i} + \hat{k}$

$\vec{b} = 2\hat{i}+\hat{j}+\hat{k}$

area of parallelogram = $|\vec{a} × \vec{b}|$

$\vec{a} \times \vec{b}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 1 & 0 & 1 \\ 2 & 1 & 1\end{array}\right|$

⇒ $\vec{a} × \vec{b} = -\hat{i}+\hat{j}+\hat{k}$

so $|\vec{a} × \vec{b}| = |\sqrt{(-1)^2+1^2+1^2}|$

$=\sqrt{3}$