Finding values that make a system of equations inconsistent I'm pretty new to matrices, and only know the basics at the moment.
I was given a question that asks:

For the system of equations:
$$
\begin{split}
x  &+ 2y &+  z &=60\\
2x &+ 3y &+  z &=85\\
3x &+ y  &+ pz &=105
\end{split}
$$
find the value of $p$ such that there is an inconsistency and hence no solutions.

I converted the system into an augmented matrix:
$$ \left[ \begin{array}{ccc|c} 1&2&1&60\\ 2&3&1&85\\ 3&1&p&105 \end{array} \right] $$
I've tried to reduce the matrix as much as I can, but I'm pretty sure I made a mistake at some point:
$$ \left[ \begin{array}{ccc|c} 1&0&-1&-10\\ 0&1&1&35\\ 0&0&(p-8)&-100 \end{array} \right] $$
How do you find the value of $p$ that makes these equations inconsistent?
And, if I did mess up the reduction, could someone please walk me through the reduction process?
Apologies if this question has been asked before, I had a good look around and didn't find anything matching this situation.
PS: I've not learnt the determinant yet.
UPDATE
As I thought, I messed up the reduction (only slightly, though). The correct reduced form is:
$$ \left[ \begin{array}{ccc|c} 1&0&-1&-10\\ 0&1&1&35\\ 0&0&(p+2)&100 \end{array} \right] $$
(Can't be reduced any further without dividing by $p$.)
For a system to be inconsistent, it must have a row of $ \left[ \begin{array}{ccc|c} 0&0&0&n\end{array} \right] $, where $n\neq0$.
Therefore, as many of you had already deduced, $p=-2$, since then $ \left[ \begin{array}{ccc|c} 0&0&0&100\end{array} \right] $.
Thanks to everyone for helping!
 A: The determinant of the coefficient for 
$$\left[ \begin{array}{ccc|c} 1&2&1&60\\ 2&3&1&85\\ 3&1&p&105 \end{array} \right]$$
is found to be $$1(3p-1)-2(2p-3)-7$$
Solve $$1(3p-1)-2(2p-3)-7=0$$ and you get $$p=-2$$
A: A definition for inconsistent system of linear equations

A system of linear equations with coefficient matrix A will be
  inconsistent for certain values on the right hand side if the row
  echelon form of A contains a row of zeros. If the row echelon form of
  the coefficient matrix A does not contain a row of zeros, then the
  system is always consistent, regardless of what the right hand side
  is.

If you calculate the row echelon for the form of the coefficients accurately, you get:
$$\begin{pmatrix}3&1&p\\ \:\:0&\frac{7}{3}&\frac{3-2p}{3}\\ \:\:0&0&\frac{p+2}{7}\end{pmatrix}$$
To get a row of zeroes (all enteries in the row must be zero), we need to have:
$$\frac{p+2}{7}=0$$
This is true for $p=-2$ only.
Hence the system is not consistent for $p=-2$.
Note that the definition is very important. I am sure your textbook has definition. Maybe you should stick to that one in your material. 
