Distance between the point and the plane

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Homework Statement


Determine the distance between the point (10,2,3) and the plane determined by the points A(2,3,0), B(3,1,4), and C(5,2,1).

The Attempt at a Solution


What I did was I took (B-A) and (C-B) and then I cross product (B-A) and (C-B) and I got -10i + 11j + 5k. After this step, I took point (10,2,3) x -10i + 11j + 5k and then I got -63. Lastly I use l-63l divide with the magnitude of the cross product which is \sqrt{(-10)^2 + 11^2 + 5^2} and I got some really weird answer which is \sqrt{246} also known as 15.68 and it end up with 63/15.68 = 4.018. I am afraid I have gone extremely wrong.
 
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ezsmith said:

Homework Statement


Determine the distance between the point (10,2,3) and the plane determined by the points A(2,3,0), B(3,1,4), and C(5,2,1).

The Attempt at a Solution


What I did was I took (B-A) and (C-B) and then I cross product (B-A) and (C-B) and I got -10i + 11j + 5k.
This is an error. The B- A= i- 2j+ 4k and C- B= 2i+ j- 3k and the cross product is
\left|\begin{array}{ccc}i & j & k \\ 1 & -2 & 4 \\ 2 & 1 & -3\end{array}\right|= (6- 4)i- (-3- 8)j+ (1+ 4)k= 2i+ 11j+ 5k
Looks like you have a sign error.

After this step, I took point (10,2,3) x -10i + 11j + 5k and then I got -63. Lastly I use l-63l divide with the magnitude of the cross product which is \sqrt{(-10)^2 + 11^2 + 5^2} and I got some really weird answer which is \sqrt{246} also known as 15.68 and it end up with 63/15.68 = 4.018. I am afraid I have gone extremely wrong.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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