We need to find for how many values of $\alpha$ the given system is inconsistent.

1. Interpret the system correctly

The equations are: [ \begin{cases} x+y+z=\alpha \ \alpha x+2\alpha y+3z=-1 \ x+3\alpha y+5z=4 \end{cases} ]

So the coefficient matrix is [ A=\begin{pmatrix} 1&1&1\ \alpha&2\alpha&3\ 1&3\alpha&5 \end{pmatrix} ] and the augmented column is [ B=\begin{pmatrix} \alpha\-1\4 \end{pmatrix}. ]

A system can be inconsistent only when the determinant of the coefficient matrix is zero.

2. Compute the determinant

[ \Delta=\begin{vmatrix} 1&1&1\ \alpha&2\alpha&3\ 1&3\alpha&5 \end{vmatrix} ]

Expand along the first row: [ \Delta=1\begin{vmatrix}2\alpha&3\3\alpha&5\end{vmatrix} -1\begin{vmatrix}\alpha&3\1&5\end{vmatrix} +1\begin{vmatrix}\alpha&2\alpha\1&3\alpha\end{vmatrix} ]

Now evaluate each minor: [ \begin{vmatrix}2\alpha&3\3\alpha&5\end{vmatrix}=10\alpha-9\alpha=\alpha ] [ \begin{vmatrix}\alpha&3\1&5\end{vmatrix}=5\alpha-3 ] [ \begin{vmatrix}\alpha&2\alpha\1&3\alpha\end{vmatrix}=3\alpha^2-2\alpha ]

Hence, [ \Delta=\alpha-(5\alpha-3)+(3\alpha^2-2\alpha) ] [ \Delta=\alpha-5\alpha+3+3\alpha^2-2\alpha ] [ \Delta=3\alpha^2-6\alpha+3 ] [ \Delta=3(\alpha^2-2\alpha+1)=3(\alpha-1)^2 ]

So, [ \Delta=0 \iff \alpha=1 ]

Thus only $\alpha=1$ can possibly make the system inconsistent.

3. Check consistency at $\alpha=1$

Substitute $\alpha=1$ into the system: [ \begin{cases} x+y+z=1 \ x+2y+3z=-1 \ x+3y+5z=4 \end{cases} ]

Now compare equations.

From second minus first: [ (x+2y+3z)-(x+y+z)=-1-1 ] [ y+2z=-2 \quad ...(1) ]

From third minus second: [ (x+3y+5z)-(x+2y+3z)=4-(-1) ] [ y+2z=5 \quad ...(2) ]

Equations (1) and (2) give [ y+2z=-2 \quad \text{and} \quad y+2z=5 ] which is impossible.

Hence the system is inconsistent for $\alpha=1$.

4. Count the number of such values

There is exactly one value of $\alpha$ for which the system is inconsistent.

So the correct option is: [ \boxed{\text{B: }1} ]

5. Compare with stored answer

Stored correct answer: B

Our derived answer: B

They agree.