If the system of linear equations : $$x+y+2z=6$$

$$2x+3y+az=a+1$$

$$-x-3y+bz=2b$$ where $$a,b \in R$$, has infinitely many solutions, then 7a + 3b is equal to :

For the system of linear equations to have infinitely many solutions, the determinant of the coefficient matrix must be zero ($$\Delta = 0$$) and the system must be consistent ($$\Delta_z = 0$$). The coefficient matrix $$A$$ is

$$A = \begin{bmatrix} 1 & 1 & 2 \\ 2 & 3 & a \\ -1 & -3 & b \end{bmatrix}$$

so its determinant is computed by expanding along the first row:

$$\Delta = 1 \cdot \det \begin{bmatrix} 3 & a \\ -3 & b \end{bmatrix} - 1 \cdot \det \begin{bmatrix} 2 & a \\ -1 & b \end{bmatrix} + 2 \cdot \det \begin{bmatrix} 2 & 3 \\ -1 & -3 \end{bmatrix}$$

Computing each $$2 \times 2$$ determinant gives:

$$\Delta = 1(3b + 3a) - 1(2b + a) + 2(-6 + 3) = 2a + b - 6$$

Setting $$\Delta = 0$$ gives $$2a + b = 6$$. Next, computing $$\Delta_z$$ by replacing the third column with the constant terms:

$$\Delta_z = \det \begin{bmatrix} 1 & 1 & 6 \\ 2 & 3 & a + 1 \\ -1 & -3 & 2b \end{bmatrix}$$

Expanding along the first row yields:

$$\Delta_z = 1 \cdot (6b + 3a + 3) - 1 \cdot (4b + a + 1) + 6 \cdot (-3) = 2a + 2b - 16$$

Setting $$\Delta_z = 0$$ gives $$2a + 2b = 16$$, which simplifies to $$a + b = 8$$.

We now have a system of two equations, $$2a + b = 6$$ and $$a + b = 8$$. Subtracting the second equation from the first gives $$a = -2$$. Substituting this value into the second equation yields $$b = 10$$.

Finally, substituting these values into the required expression:

$$7a + 3b = 7(-2) + 3(10) = -14 + 30 = 16$$

Therefore, the final answer is $$16$$.