How to get the value of t For each number t, consider the triangle T with corners at, A = (1, 2, 3), B = (1, 0, −2),
C = (t, t, 1).
For which value of t does T have a right angle at A?
I know how it has to do with AB*CD = something
Rest i don't know how to solve this??

As I commented earlier, you can use the Pythagoras theorem: for a right angle we know that $a^2+b^2=c^2$. We can find the lengths first:
$AB = \sqrt{0^2+2^2+5^2} = \sqrt{29}$
$BC = \sqrt{(1-t)^2 + t^2 +3^2} = \sqrt{10-2t+2t^2}$
$AC = \sqrt{(1-t)^2 + (2-t)^2 + 2^2} = \sqrt{9-6t+2t^2}$
Now we can plug in these values for $a,b$ and $c$. We want a right angle at $A$ so we need to solve: $(AB)^2 + (AC)^2 = (BC)^2$.
This gives: $t=7$.
