1. For a system of 3 linear equations in 3 variables to have infinitely many solutions, the equations must be dependent.

   That means one equation should be a linear combination of the other two, and hence the coefficient matrix and augmented matrix must both have rank less than 3.

2. The given system is:

   $$\begin{aligned} x+y+z&=6 \quad \cdots (1)\\ 2x+5y+\alpha z&=\beta \quad \cdots (2)\\ x+2y+3z&=14 \quad \cdots (3) \end{aligned}$$

3. Since equation (2) must be dependent on equations (1) and (3), let

   $$(2)=m(1)+n(3).$$

   Then comparing coefficients:

   • Coefficient of $x$: $m+n=2$
   • Coefficient of $y$: $m+2n=5$

4. Solve these two equations:

   $$(m+2n)-(m+n)=5-2$$

   $n=3$ Then $m+3=2 \implies m=-1.$

5. Now compare the $z$-coefficients:

   $$\alpha = m(1)+n(3)=(-1)(1)+(3)(3)=-1+9=8.$$

6. Compare the constants on the right-hand side:

   $$\beta = m(6)+n(14)=(-1)(6)+(3)(14)=-6+42=36.$$

7. Therefore,

   $$\alpha+\beta=8+36=44.$$

8. Checking options:

   • A: $8$
   • B: $36$
   • C: $44$ ✅
   • D: $48$

So the correct answer is Option C.