What is the height of the pyramid with these four points? I have a pyramid, whose base is formed of points $A (3,5,3) $, $B (-2,11,-5) $, $C (1,-1,4) $ . Now I need to find the height of the pyramid from the point $S (0,6,4) $ so my idea was this. I form a plain using vectors $AB $ and $AC$, then I find a vector which is orthagonal to that plain. After that I found the intersection of the orthagonal vector and the plain, let that be point $T $. Then my height is actually the length of vector $ST$. Here I get that the length is 63, but my textbook says it's 3.. just wondering if I made a mistake somewhere or if the solution is wrong. Thanks
 A: Your approach is perfect.  If you show your calculations, we can help figure out where you went wrong.  One way of finding that orthogonal vector would be to find two vectors in the plane, then take their cross product.  For example, if you draw a picture, you'll notice that vectors $A-C$ and $A-B$ lie in that plane (i.e. what you're calling $AB$ and $AC$).  Computing $(A-C) \times (A-B)$ yields an orthogonal vector. 

Another approach that is overly cumbersome in practice but of pedagogical value is as follows:
If one knows only the side lengths of a triangle, Heron's formula can be used to compute its area.  A generalization of this formula exists that computes the volume of a tetrahedron given the lengths of its sides.  We can find all the side lengths of the tetrahedron in question via repeated applications of the Pythagorean theorem.  This gives us all the necessary information to compute the area of the pyramid's triangular base as well as the volume of the entire tetrahedron.
Recall that the volume of a tetrahedron is given by $\displaystyle V = \frac{1}{3}bh$, where $b$ is the area of its base.  Heron's formula reveals the values of $b$ and $V$, allowing us to solve for $h$.
A: The given problem is equivalent to finding the distance from the origin for the plane $\pi$ through
$$ A'(3,-1,-1),\qquad B'(-2,5,-9),\qquad C'(1,-7,0) $$
whose equation is given (by solving a $3\times 3$ system) by
$$ 2x-y-2z=9. $$
The distance of $\pi$ from the origin is so given by
$$ \frac{9}{\sqrt{2^2+1^2+2^2}}=\frac{9}{3}=\color{red}{3}.$$

$63$ is clearly way too much since
$$ d(S,\pi_{ABC}) \leq d(S,A) = d(O,A')=\sqrt{11}.$$
A: $$\vec{AB}=(-5,6,-8)~ ;~ \vec{AC}=(-2,-6,1)$$
We can use cross product to find that $\vec N = (-2,1,2)$ is a normal to $\vec{AB} , \vec {AC}$. Thus, the plain equation is:
$$-2x+y+2z-5=0$$
The distance from a given point $p(x_0 , y_0 , z_0)$ to the plain is given by:
$$dist(p)=\frac{|-2 \cdot x_0+ 1 \cdot y_0 + 2 \cdot z_0 -5|}{\sqrt{(-2)^2+1^2+2^2}}=\frac{|-2 \cdot x_0+ 1 \cdot y_0 + 2 \cdot z_0 -5|}{3}$$
Setting $p=S=(0,6,4)$ yields:
$$\frac{|-2\cdot 0 + 1 \cdot 6 + 2 \cdot 4 -5|}{\sqrt{(-2)^2+1^2+2^2}}=3$$
