Answer
$\lt 0,2,2 \gt$ or, $2j+2k$
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 $u \times v$.
Now, $u \times v=\lt (1)(0)-(-1)(0),(-1)(-2)-(1)(0), (1)(0) -(1)(-2) \gt=\lt 0,2,2 \gt$
or, $2j+2k$
