Finding an area of a triangle formed by three points

[CrosisBH]

Homework Statement

P(3, 0, 3), Q(−2, 1, 5), R(6, 2, 7)
(a) Find a nonzero vector orthogonal to the plane through the points P, Q,
and R.

(b) Find the area of the triangle PQR

Homework Equations

A = \frac{1}{2}|\vec{AB}\times\vec{AC}|
Source: http://www.maths.usyd.edu.au/u/MOW/vectors/vectors-11/v-11-7.html

The Attempt at a Solution

The first part was straight forward and my homework marked it right:

P(3,0,3)\\<br /> Q(-2,1,5)\\<br /> R(6,2,7)\\<br /> \vec{PR} = \langle 3,2,4 \rangle\\<br /> \vec{PQ} = \langle 5,1,2 \rangle
To find a vector orthogonal to two vectors, you do a cross product:
\vec{PR} \times \vec {PQ} = \langle 0,14,-7 \rangle
My homework marked this part right for this answer.

Part two:
First taking the magnitude of the cross product:
|\vec{PR} \times \vec {PQ}| = \sqrt{245} = 7\sqrt{5}
Dividing it by two:
A = \frac{7}{2}\sqrt{5}
Which is my final answer, and the website marks it as wrong, but I can't see why it's wrong. The magnitude is right which I triple checked, by my own hands and Wolfram Alpha, and the order of the cross product is irrelevant in this situation.

I found a calculator that calculates the area from three 3D points and it spit out the nonexact answer of 14.534. My answer is approximately 7.826, a little more than half. I can't see my error unless I'm completely calculating this area wrong. Thanks for the help.

 

[fresh_42]

$\vec {PQ}$ looks wrong.

 

[CrosisBH]

$\vec {PQ}$ looks wrong.

AHHH it was that simple. I was successful in finding a orthogonal vector, but it wasn't the cross product of the two supposed vectors from the points.

Correction:
\vec{PQ} = \langle -5, 1, 2 \rangle
Making the normal vector of interest
\vec{PR} \times \vec {PQ} = \langle 0,-26,13 \rangle
|\vec{PR} \times \vec {PQ}| = 13\sqrt{5}
So the answer is
\frac{13\sqrt{5}}{2}

Let it be the simple subtraction that kills me haha. Thank you