Q: Area of triangle determined by three vectors in $\mathbb{R}^3$ Given vectors $a = (2, 1, 3)$, $b = (4, 1, 2)$, $c = (1, -1, 5)$, need to find the area of the triangle $abc$ determined by the three vectors (the vectors are the vertices of the triangle).
My solution: 
$A = 1/2 \text{(base)(height)}$. Take base $d := a-b = (-2, 0, 1)$. Need to find the shortest (perpendicular) distance from $c$ to the line determined by $L := a + td$ ($t\in \mathbb{R}$). $L = (2,1,3)+t(-2,0,1)$, thus we need to find $x = (2-2t,1, 3+t)$. We need $x\cdot d = 0$, so $t=1/5$ and $x=(3/5,1,16/5)$. We have that height $h:=c-x=(2/5,-2,9/5)$, so $\vert h \vert = \sqrt{37/5}$, and also $\vert d \vert = \sqrt{5}$. Hence, the area is $A = 1/2 * 37/5 * \sqrt{5}=\frac{37}{{2\sqrt{5}}}$.
Please let me know if you think my solution is correct. I'd like to make sure that I didn't make some conceptual mistake(s).

A: I would've gone with the cross product approach suggested in the comments (see below), but here's how I'd do it using your approach.
By taking the base of the triangle as $d=a-b$, we're effectively translating the origin to $b$. To find the altitude, we need the component of $e=c-b$ that is orthogonal to $d$, i.e., the orthogonal rejection of $e$ relative to $d$. The orthogonal projection of $e$ onto $d$ is $${e\cdot d\over d\cdot d}d=\frac9 5(-2,0,1)=\left(\frac{18}5,0,\frac95\right),$$ and the orthogonal rejection is $e$ minus this vector, i.e., $\left(\frac35,-2,\frac65\right)$. The area of the triangle is then $$A=\frac12\|(-2,0,1)\|\cdot\left\|\left(\frac35,-2,\frac65\right)\right\|=\frac12\cdot\sqrt{5}\cdot{\sqrt{145}\over5}={\sqrt{29}\over2}.$$ You can also find the altitude by finding the nearest point on the line $a+td$ to $c$, as you do, by solving $(c-a-td)\cdot d=0$ for $t$. You'll find that when you substitute the value of $t$ that you get back into $c-a-td$, you'll end up with the same vector as above.  
Alternatively, the norm of the cross product of two vectors is the area of the parallelogram defined by those vectors, so we have for the triangle $$A=\frac12\|d\times e\| = \frac12\|(2,3,4)\|={\sqrt{29}\over 2}.$$

A: Another method is to use one of the trigonometric versions for the area of a triangle , $ \ A \ = \ \frac{1}{2} \ a \ b \ \sin \theta \ $ , where $ \ \theta \ $ is the included angle between the given sides.  You can pick any two of the vector lengths to use for $ \ a \ $ and $ \ b \ $ , say , $ \ \langle \ 2, \ 0, \ -1 \ \rangle \ $ (with length $ \ \sqrt{5} \ $ ) and $ \ \langle \ -1  , \ -2 , \ 2 \ \rangle \ $ (with length $ \ 3 \ $ ) .  The cosine of the included angle is given by $ \ \frac{\vec{a} \ \cdot \ \vec{b} }{a \ b } \ $ , so the "Pythagorean Identity" gives the sine of this angle as $$ \ \frac{\sqrt{a^2b^2 \ - \ (\vec{a} \ \cdot \vec{b})^2}}{a \ b} \ \ ;  $$
the area of the triangle is then $ \ A \ = \ \frac{1}{2} \ \sqrt{a^2b^2 \ - \ (\vec{a} \ \cdot \vec{b})^2} \ $ .  We find the dot product of our chosen vectors to be $ \ -4 \ $ (the triangle is obtuse) , and the area as 
$$ \ A \ = \ \frac{1}{2} \ \sqrt{(\sqrt{5})^2  \cdot 3^2 \ - \ (-4)^2} \ \ = \ \ \frac{1}{2} \ \sqrt{5 \cdot 9 \ - \ 16} \ \ = \ \ \frac{1}{2} \ \sqrt{29} \ \ . $$

A: You could use Heron's Formula? https://en.wikipedia.org/wiki/Heron%27s_formula

Convert the vertex vectors into three edge vectors. Then compute length of each edge. Use that to find the semi-perimeter. Plug in into the formula. 
I find it easiest to use vector manipulation when the co-ordinates are non-planar. 
a<-c(2,1,3)
b<-c(4,1,2)
c<-c(1,-1,5)

s1<-b-a
s2<-c-b
s3<-a-c

s1.length<-norm(s1,type="2")
s2.length<-norm(s2,type="2")
s3.length<-norm(s3,type="2")

p<-(s1.length+s2.length+s3.length)/2
sqrt(p*(p-s1.length)*(p-s2.length)*(p-s3.length))

I get Area as 2.692582. 
