If the system of equations
3x + y + 4z = 3
$$2x+\alpha y-z = -3$$
x+ 2y + z = 4
has no solution, then the value of $$\alpha$$ is equal to :

For the system to have no solution, we need the determinant of the coefficient matrix to be zero and the system to be inconsistent.

The determinant of the coefficient matrix is given by$$D = \begin{vmatrix} 3 & 1 & 4 \\ 2 & \alpha & -1 \\ 1 & 2 & 1 \end{vmatrix}$$

Expanding, we get $$D = 3(\alpha + 2) - 1(2 + 1) + 4(4 - \alpha) = 3\alpha + 6 - 3 + 16 - 4\alpha = -\alpha + 19$$.

For no solution: $$D = 0 \Rightarrow \alpha = 19$$.

With $$\alpha = 19$$, equation 2 becomes: $$2x + 19y - z = -3$$.

Subtracting 3 times equation 3 from equation 1 gives $$0 - 5y + z = -9$$, so $$-5y + z = -9$$.

Subtracting 2 times equation 3 from equation 2 gives $$0 + 15y - 3z = -11$$, so $$15y - 3z = -11$$.

From these, $$3(-5y + z) = -27 \Rightarrow -15y + 3z = -27$$.

Adding this to $$15y - 3z = -11$$ gives $$0 = -38 \neq 0$$, so the system is inconsistent. ✓

The answer is Option 1: 19.