The set of equations $x + y + z = 1$, $ax – ay + 3z = 5$, and $5x – 3y + az = 6$ has infinite solution if $a = 3$.
To see this, we can first express the equations in terms of $x$, $y$, and $z$:
\begin{align}
x &= 1 – y – z \
ax – ay + 3z &= 5 \
5x – 3y + az &= 6
\end{align}
Subtracting the second equation from the third equation, we get:
\begin{align}
4x – 2y &= 1 \
\Rightarrow x – \frac{y}{2} &= \frac{1}{4}
\end{align}
Substituting this into the first equation, we get:
\begin{align}
1 – y – z &= 1 – \frac{y}{2} – z \
\Rightarrow z &= \frac{1}{2}
\end{align}
Substituting this into the second equation, we get:
\begin{align}
3 \left( \frac{1}{2} \right) – ay + 3z &= 5 \
\Rightarrow -ay + 3 &= 5 \
\Rightarrow -ay &= 2 \
\Rightarrow y &= -\frac{2}{a}
\end{align}
Therefore, the solution to the system of equations is:
\begin{align}
x &= \frac{1}{4} \
y &= -\frac{2}{a} \
z &= \frac{1}{2}
\end{align}
Note that this solution is valid for any value of $a$, including $a = 3$. Therefore, the set of equations has infinite solution if $a = 3$.
For the other options, $a = -4$, $a = -3$, and $a = 4$, the system of equations has no solution. This can be seen by substituting these values of $a$ into the equations and trying to solve for $x$, $y$, and $z$. In each case, we will get an equation that is inconsistent with the other two equations.