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 and the system must be consistent. 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: $$ \det(A) = 1 \cdot \det \begin{bmatrix} 3 & a \\ -3 & b \end{bmatrix} - 1 \cdot \det \begin{bmatrix} 2 & a \\ -1 & b \end{bmatrix} + 2 \cdot \det \begin{bmatrix} 2 & 3 \\ -1 & -3 \end{bmatrix}. $$

Computing each 2×2 determinant gives $$ \det \begin{bmatrix} 3 & a \\ -3 & b \end{bmatrix} = 3b - (-3a) = 3a + 3b, \quad \det \begin{bmatrix} 2 & a \\ -1 & b \end{bmatrix} = 2b - (-a) = 2b + a, \quad \det \begin{bmatrix} 2 & 3 \\ -1 & -3 \end{bmatrix} = 2 \cdot (-3) - 3 \cdot (-1) = -6 + 3 = -3. $$ Substituting these results into the determinant expression yields $$ \det(A) = 1 \cdot (3a + 3b) - 1 \cdot (2b + a) + 2 \cdot (-3) = 3a + 3b - 2b - a - 6 = 2a + b - 6, $$ and setting $$\det(A) = 0$$ for infinitely many solutions gives $$ 2a + b - 6 = 0 \quad\Rightarrow\quad 2a + b = 6, $$ so $$b = 6 - 2a$$.

Substituting $$b = 6 - 2a$$ into the original system $$ x + y + 2z = 6, \quad 2x + 3y + az = a + 1, \quad -\,x - 3y + (6 - 2a)z = 12 - 4a $$ and eliminating variables in terms of $$z$$ leads to infinitely many solutions. Multiplying the first equation by 2 gives $$2x + 2y + 4z = 12$$, so subtracting this from the second yields $$ (2x + 3y + az) - (2x + 2y + 4z) = (a + 1) - 12, $$ which simplifies to $$ y + (a - 4)z = a - 11, \quad y = (a - 11) - (a - 4)z. $$ Substituting this into the first equation then gives $$ x + \bigl[(a - 11) - (a - 4)z\bigr] + 2z = 6 \quad\Rightarrow\quad x = 17 - a + (a - 6)z. $$

Replacing both $$x$$ and $$y$$ in the third equation $$ -\bigl[17 - a + (a - 6)z\bigr] - 3\bigl[(a - 11) - (a - 4)z\bigr] + (6 - 2a)z = 12 - 4a $$ and expanding gives $$ -17 + a - (a - 6)z - 3(a - 11) + 3(a - 4)z + (6 - 2a)z = 12 - 4a. $$ Combining constant terms and the coefficients of $$z$$ shows $$ (-17 + 33) + (a - 3a) = 16 - 2a, \quad -(a - 6) + (3a - 12) + (6 - 2a) = 0, $$ so the equation reduces to $$ 16 - 2a = 12 - 4a, \quad 2a = -4 \quad\Rightarrow\quad a = -2. $$

Substituting $$a = -2$$ into $$2a + b = 6$$ yields $$ -4 + b = 6 \quad\Rightarrow\quad b = 10. $$ Then $$ 7a + 3b = 7(-2) + 3(10) = -14 + 30 = 16. $$ Verification with $$a = -2$$ and $$b = 10$$ confirms that the system is indeed consistent and dependent, so infinitely many solutions exist. Therefore, the final answer is $$7a + 3b = 16$$.