Answer
$12 \sqrt {10}$
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 $\triangle PQR=\dfrac{1}{2}| \overrightarrow {PQ} \times \overrightarrow {PR}|$
$\overrightarrow {PQ} \times \overrightarrow {PR}=\begin{vmatrix}i&j&k\\-4&-2&-12\\0&6&0 \end{vmatrix}=72i--24k$
Then, we have
$A=|\overrightarrow {PQ} \times \overrightarrow {PR}|=\sqrt {(72)^2+(-24)^2}=\sqrt{5760}=24 \sqrt {10}$
Thus, the area of Area of $\triangle PQR$ is $\dfrac{24 \sqrt {10}}{2}=12 \sqrt {10}$
