The set of equations

                                       x+y+z=1

                                       ax-ay +3z =5

                                       5x—3y+az =6

has infinite solutions, if a =

The correct answer is 4.

We are given the following system of linear equations:
1) $x+y+z=1$
2) $ax-ay+3z=5$
3) $5x-3y+az=6$

For a system of linear equations to have infinitely many solutions, the determinant of the coefficient matrix, denoted by $\Delta$ (or D), must be equal to zero.

Let's write down the determinant of the coefficient matrix:
$\Delta =\left|\begin{array}{ccc}1& 1& 1\\ a& -a& 3\\ 5& -3& a\end{array}\right|$

We expand this determinant along the first row:
$\Delta =1·(-a·a-3·(-3\left)\right)-1·(a·a-3·5)+1·(a·(-3)-(-a)·5)$
$\Delta =1·(-a^2+9)-1·(a^2-15)+1·(-3a+5a)$
$\Delta =-a^2+9-a^2+15+2a$
$\Delta =-2a^2+2a+24$

Setting $\Delta =0$ for infinite solutions:
$-2a^2+2a+24=0$
Dividing the entire equation by -2:
$a^2-a-12=0$

Now, we factor the quadratic equation:
$(a-4)(a+3)=0$
This gives two possible values for a:
$a=4$ or $a=-3$

Let's verify these values with the other determinants to ensure the system is consistent (which is required for infinite solutions, rather than no solutions).

For $a=4$, the equations become:
1) $x+y+z=1$
2) $4x-4y+3z=5$
3) $5x-3y+4z=6$

If we add equation 1 and equation 2:
$(x+y+z)+(4x-4y+3z)=1+5$
$5x-3y+4z=6$
This is exactly equation 3. Since one of the equations is a linear combination of the other two, the system is consistent and has infinitely many solutions. Thus, $a=4$ yields infinite solutions.