Using vectors, find the area of the triangle with vertices: $A(1,2,3),B(2,-1,4)$ and $C(4,5,-1)$.

Hint: First we have to find out the sides $\overrightarrow{AB}$ and $\overrightarrow{AC}$. $\overrightarrow{AB}$ is found by finding $\overrightarrow{AB}=B-A$ and $\overrightarrow{AC}$ is found by finding $\overrightarrow{AC}=C-A$. Then we can find $\overrightarrow{AB}\times \overrightarrow{BC}$ and use formula - Area$=\dfrac{1}{2}\overrightarrow{AB}\times \overrightarrow{BC}$ to find the area of the given triangle.

Complete Step-by-step answer:
Before proceeding with the question, we must be familiar with the values of cross-products, like $\widehat{i}\times \widehat{j}=\widehat{k}$ , $\widehat{i}\times \widehat{i}=0,\widehat{i}\times \widehat{k}=-\widehat{j},\widehat{j}\times \widehat{i}=-\widehat{k},\widehat{j}\times \widehat{j}=0,\widehat{j}\times \widehat{k}=\widehat{i},\widehat{k}\times \widehat{i}=\widehat{j},\widehat{k}\times \widehat{j}=-\widehat{i},\widehat{k}\times \widehat{k}=0$like we must know the formula for finding the area of a triangle using vectors. The area of the triangle is given by $\dfrac{1}{2}\overrightarrow{AB}\times \overrightarrow{BC}$.
In this question, we have to find the area of the triangle, where the vertices of the triangle are $A(1,2,3),B(2,-1,4)$ and $C(4,5,-1)$. It can be represented as shown below,

First, we have to find out $\overrightarrow{AB}$ and $\overrightarrow{AC}$.
$\overrightarrow{AB}$ is found by finding $\overrightarrow{AB}=B-A$, where $A(1,2,3),B(2,-1,4)$.
$\Rightarrow \overrightarrow{AB}=\left( 2-1 \right)\widehat{i}+\left( -1-2 \right)\widehat{j}+\left( 4-3 \right)\widehat{k}$
$\therefore \overrightarrow{AB}=\widehat{i}-3\widehat{j}+\widehat{k}$
$\overrightarrow{AC}$ found by finding $\overrightarrow{AC}=C-A$, where $C(4,5,-1)$ and $A(1,2,3)$.
$\Rightarrow \overrightarrow{AC}=\left( 4-1 \right)\widehat{i}+\left( 5-2 \right)\widehat{j}+\left( \left( -1 \right)-3 \right)$
$\therefore \overrightarrow{AC}=3\widehat{i}+3\widehat{j}-4\widehat{k}$
Now we have to find $\overrightarrow{AB}\times \overrightarrow{BC}$.
$\overrightarrow{AB}\times \overrightarrow{BC}=(\widehat{i}-3\widehat{j}+\widehat{k})\times (3\widehat{i}+3\widehat{j}-4\widehat{k})$
Cross product of $\widehat{i}\times \widehat{j}=\widehat{k}$ , $\widehat{i}\times \widehat{i}=0,\widehat{i}\times \widehat{k}=-\widehat{j},\widehat{j}\times \widehat{i}=-\widehat{k},\widehat{j}\times \widehat{j}=0,\widehat{j}\times \widehat{k}=\widehat{i},\widehat{k}\times \widehat{i}=\widehat{j},\widehat{k}\times \widehat{j}=-\widehat{i},\widehat{k}\times \widehat{k}=0$
Opening the brackets we get:
$\Rightarrow \widehat{i}(12-3)-\widehat{j}(-4-3)+\widehat{k}(3+9)$
$\Rightarrow \overrightarrow{AB}\times \overrightarrow{BC}=9\widehat{i}+7\widehat{j}+12\widehat{k}$
$\left| \overrightarrow{AB}\times \overrightarrow{BC} \right|=$is given by $\sqrt{{{9}^{2}}+{{7}^{2}}+{{12}^{2}}}=\sqrt{274}$
Therefore, the area of the triangle ABC is given by $\dfrac{1}{2}\left| \overrightarrow{AB}\times \overrightarrow{BC} \right|=\dfrac{\sqrt{274}}{2}$
Hence, the area of the triangle is given by $\dfrac{\sqrt{274}}{2}$sq.units.

Note: Be careful while calculating the side of the triangle $\overrightarrow{AB}$ as it can only be found by $\overrightarrow{AB}=B-A$ so do not take it as $A-B$. Always remember that the formula for the area of the triangle is enclosed under modulus so do not forget the modulus while applying the formula. If the found value of the area is negative then the answer is wrong because the value of the area can never be negative.