Answer
$\lt 12,43,10 \gt$
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 through the points P,Q,R can be written as $u \times v$.
Now, $u \times v=\lt (2)(6)-(5)(0),(-5)(-5)-(-3)(6), (-3)(0) -(2)(-5) \gt=\lt 12,43,10 \gt$
