If the system of linear equations :

$$\begin{aligned} & x+y+2 z=6 \\ & 2 x+3 y+\mathrm{az}=\mathrm{a}+1 \\ & -x-3 y+\mathrm{b} z=2 \mathrm{~b} \end{aligned}$$

where $a, b \in \mathbf{R}$, has infinitely many solutions, then $7 a+3 b$ is equal to :

<p>We begin with the system:</p> <p>$$ \begin{aligned} x+y+2z &= 6, \\ 2x+3y+az &= a+1, \\ -x-3y+bz &= 2b. \end{aligned} $$</p> <p><strong>Step 1.</strong> Solve the first equation for $x$:</p> <p>$x = 6 - y - 2z.$</p> <p><strong>Step 2.</strong> Substitute $x = 6-y-2z$ into the second equation:</p> <p>$2(6-y-2z) + 3y + az = a+1.$</p> <p>Expanding and simplifying:</p> <p>$12 - 2y - 4z + 3y + az = a+1 \quad \Longrightarrow \quad y + (a-4)z = a - 11.$</p> <p>Call this <strong>Equation (I)</strong>.</p> <p><strong>Step 3.</strong> Substitute $x = 6-y-2z$ into the third equation:</p> <p>$-(6-y-2z) - 3y + bz = 2b.$</p> <p>Expanding and simplifying:</p> <p>$-6 + y + 2z - 3y + bz = 2b \quad \Longrightarrow \quad -2y + (b+2)z = 2b + 6.$</p> <p>Call this <strong>Equation (II)</strong>.</p> <p><strong>Step 4.</strong> For the system to have infinitely many solutions, the two equations in $y$ and $z$ must be dependent—that is, one must be a constant multiple of the other. Assume there exists a constant $k$ such that</p> <p>$-2 = k \cdot 1 \quad \Longrightarrow \quad k = -2.$</p> <p>Apply this to the coefficient of $z$ and the constant term.</p> <p>For the $z$-coefficient in Equations (I) and (II):</p> <p>$b+2 = k(a-4) = -2(a-4) = -2a+8.$</p> <p>Thus,</p> <p>$b = -2a+6.$</p> <p>For the constant term:</p> <p>$2b+6 = k(a-11) = -2(a-11) = -2a + 22.$</p> <p>Substitute $b = -2a+6$ into this equation:</p> <p>$2(-2a+6) + 6 = -2a + 22 \quad \Longrightarrow \quad -4a + 12 + 6 = -2a + 22.$</p> <p>Simplify:</p> <p>$-4a + 18 = -2a + 22.$</p> <p>Solve for $a$:</p> <p>$-4a + 18 + 4a = -2a + 22 + 4a \quad \Longrightarrow \quad 18 = 2a + 22,$</p> <p>$2a = 18 - 22 = -4 \quad \Longrightarrow \quad a = -2.$</p> <p>Substitute $a = -2$ into $b = -2a+6$:</p> <p>$b = -2(-2) + 6 = 4 + 6 = 10.$</p> <p><strong>Step 5.</strong> We now compute</p> <p>$7a + 3b = 7(-2) + 3(10) = -14 + 30 = 16.$</p> <p>Thus, the value of $7a+3b$ is</p> <p>$\boxed{16}.$</p>