The area of a triangle formed by vectors P = (3 i - 4 j + 2 k ) m...

The area of a triangle with adjacent sides $\vec{P}$ and $\vec{Q}$ is given by $\dfrac{1}{2} |\vec{P} \times \vec{Q}|$.

Calculating the cross product of the given vectors:

$\vec{P} \times \vec{Q} = \begin{vmatrix} \hat{i} \text{ and } \hat{j} \text{ and } \hat{k} \\ 3 \text{ and } -4 \text{ and } 2 \\ 1 \text{ and } 1 \text{ and } -2 \end{vmatrix}$

$\vec{P} \times \vec{Q} = \hat{i}(8 - 2) - \hat{j}(-6 - 2) + \hat{k}(3 + 4) = 6\hat{i} + 8\hat{j} + 7\hat{k}$

Finding the magnitude of the cross product:

$|\vec{P} \times \vec{Q}| = \sqrt{6^2 + 8^2 + 7^2} = \sqrt{36 + 64 + 49} = \sqrt{149}$

The area of the triangle is:

$\text{Area} = \dfrac{1}{2} \sqrt{149} \approx \dfrac{12.2}{2} = 6.1$

Approximating to the nearest integer, the numerical value of the area is $6$.

Answer: $6\ \text{m}^2$