Answer
$\dfrac{\sqrt{11}}{2}$
Work Step by Step
The cross product is defined as:
$u \times v=\begin{vmatrix}i&j&k\\m_1&m_2&m_3\\n_1&n_2&n_3\end{vmatrix}=\lt m_2n_3-m_3n_2, m_3n_1-m_1n_3, m_1n_2-m_2b_1 \gt$
Let us consider $u=\overrightarrow {PQ}$ and $v=\overrightarrow {PR}$
Then, the vector perpendicular to the plane passing throght the points P,Q,R can be written as $u \times v$.
Area of $\triangle PQR=\dfrac{1}{2}| \overrightarrow {PQ} \times \overrightarrow {PR}|$
$\overrightarrow {PQ} \times \overrightarrow {PR}=\begin{vmatrix}i&j&k\\-2&-1&-1\\-1&1&0 \end{vmatrix}=i+j-3k$
Then, we have
$A=|\overrightarrow {PQ} \times \overrightarrow {PR}|=\sqrt {(1)^2+(1)^2+(-3)^2}=\sqrt{11}$
Thus, the area of parallelogram $=\dfrac{\sqrt{11}}{2}$
