Answer
$\sqrt {14}$
Work Step by Step
Step 1. Establish two vectors of the triangle: $\vec {PQ}=\langle 0-1, 1-0, 0-1 \rangle=\langle -1, 1, -1 \rangle$ and $\vec {PR}=\langle 2-1, 3-0, 4-1 \rangle=\langle 1, 3, 3 \rangle$
Step 2. Calculate the length of the cross product of the two vectors: $\vec {PQ}\times\vec {PR}
=\langle (1\times3+1\times3), (-1\times1+1\times3), (-1\times3-1\times1) \rangle=\langle 6, 2, -4 \rangle$, and $|\vec {PQ}\times\vec {PR}|=\sqrt {6^2+2^2+(-4)^2}=\sqrt {56}=2\sqrt {14}$
Step 3. The area of the triangle is half of the parallelogram formed by the two vectors. We know the area of this parallelogram from step 2, thus the area of the triangle is $\sqrt {14}$
