Let $$[\lambda]$$ be the greatest integer less than or equal to $$\lambda$$. The set of all values of $$\lambda$$ for which the system of linear equations $$x + y + z = 4$$, $$3x + 2y + 5z = 3$$, $$9x + 4y + (28 + [\lambda])z = [\lambda]$$ has a solution is:

Let us denote the greatest-integer function by a square bracket. Hence we write $$[\lambda]=a,$$ where $$a$$ is always an integer and satisfies $$a\le \lambda<a+1.$$

The given system of equations can, after this substitution, be rewritten as

$$\begin{aligned} x+y+z &=4,\\ 3x+2y+5z &=3,\\ 9x+4y+(28+a)z &=a. \end{aligned}$$

To investigate the existence of solutions, we first look at the determinant of the coefficient matrix. Stating the formula, for a $$3\times3$$ matrix $$\begin{vmatrix} p_{11}&p_{12}&p_{13}\\ p_{21}&p_{22}&p_{23}\\ p_{31}&p_{32}&p_{33} \end{vmatrix}=p_{11}(p_{22}p_{33}-p_{23}p_{32})-p_{12}(p_{21}p_{33}-p_{23}p_{31})+p_{13}(p_{21}p_{32}-p_{22}p_{31}).$$ We now substitute $$\begin{pmatrix} p_{11}&p_{12}&p_{13}\\ p_{21}&p_{22}&p_{23}\\ p_{31}&p_{32}&p_{33} \end{pmatrix}= \begin{pmatrix} 1&1&1\\ 3&2&5\\ 9&4&28+a \end{pmatrix}.$$

Using the formula term by term, we have

$$\begin{aligned} \Delta&=1\bigl(2(28+a)-5\cdot4\bigr) -1\bigl(3(28+a)-5\cdot9\bigr) +1\bigl(3\cdot4-2\cdot9\bigr)\\[4pt] &=1\bigl(56+2a-20\bigr) -\bigl(84+3a-45\bigr) +\bigl(12-18\bigr)\\[4pt] &=\bigl(36+2a\bigr)-\bigl(39+3a\bigr)-6\\[4pt] &=36+2a-39-3a-6\\[4pt] &=-9-a. \end{aligned}$$

So the determinant of the coefficient matrix is $$\Delta=-(a+9).$$ Two distinct possibilities arise:

1. $$a\neq-9$$

If $$a\neq-9$$, then $$\Delta\neq0.$$ Because the determinant is non-zero, the coefficient matrix is non-singular, and by the theory of linear equations a unique solution exists. Hence the system is consistent for every integer $$a$$ except $$a=-9$$, or in other words, for all $$\lambda$$ whose greatest integer is not $$-9.$$

2. $$a=-9$$

Now we put $$a=-9$$ directly into the system:

$$\begin{aligned} x+y+z &=4,\\ 3x+2y+5z &=3,\\ 9x+4y+19z &=-9. \end{aligned}$$

Because the determinant has become zero, we must compare the ranks of the coefficient matrix and the augmented matrix. A straightforward way is to see whether the third equation can be derived from the first two.

Multiply the second equation by $$3$$ to match the coefficient of $$x$$ in the third equation:

$$3(3x+2y+5z)=9x+6y+15z=9.$$

Subtract this new equation from the third equation:

$$\bigl(9x+4y+19z\bigr)-\bigl(9x+6y+15z\bigr)=(4y-6y)+(19z-15z)=-2y+4z,$$ and on the right hand side $$-9-9=-18.$$ So we receive the relation $$-2y+4z=-18.$$

At the same time, from the first equation $$y=4-x-z.$$ Substituting into $$-2y+4z=-18$$ gives $$-2(4-x-z)+4z=-18.$$ This simplifies step by step: $$-8+2x+2z+4z=-18,$$ $$2x+6z-8=-18,$$ $$2x+6z=-10,$$ $$x+3z=-5.$$

Now look back at the second original equation with $$a=-9$$, i.e. $$3x+2y+5z=3.$$ Replacing $$y$$ by $$4-x-z:$$ $$3x+2(4-x-z)+5z=3,$$ $$3x+8-2x-2z+5z=3,$$ $$x+3z+8=3,$$ $$x+3z=-5.$$

This is exactly the same condition we obtained from the difference of equations, hence it is automatically satisfied. Therefore the third equation is a linear combination of the first two, the ranks of both the coefficient and augmented matrices are equal (both equal to $$2$$), and the system has infinitely many solutions. Hence it is still consistent.

Since the system is consistent for $$a=-9$$ as well, there is no value of $$a$$ (and thus no value of $$\lambda$$) that makes the system inconsistent.

Because $$a=[\lambda]$$ can be any integer and we have shown consistency for every possible integer, the original parameter $$\lambda$$ can be any real number. Symbolically, the admissible set is $$\mathbb R.$$

Hence, the correct answer is Option A.