For the system of equations

$x+y+z=6$

$x+2 y+\alpha z=10$

$x+3 y+5 z=\beta$, which one of the following is NOT true?

Given system of equations, <br/><br/>$$ \begin{aligned} x+y+z & =6 ........(i)\\\\ x+2 y+\alpha z & =10 ........(ii)\\\\ x+3 y+5 z & =\beta ........(iii) \end{aligned} $$ <br/><br/>Here, <br/><br/>$$ \begin{aligned} \Delta & =\left|\begin{array}{lll} 1 & 1 & 1 \\ 1 & 2 & \alpha \\ 1 & 3 & 5 \end{array}\right| \\\\ & =1(10-3 \alpha)-1(5-\alpha)+1(3-2) \\\\ & =10-3 \alpha-5+\alpha+1 \\\\ & =6-2 \alpha \end{aligned} $$ <br/><br/>For unique solution, $\Delta \neq 0$ <br/><br/>$\Rightarrow 6-2 \alpha \neq 0 \Rightarrow \alpha \neq 3$ <br/><br/>When, $\alpha=3$ <br/><br/>$$ \begin{aligned} \Delta_1 & =\left|\begin{array}{ccc} 6 & 1 & 1 \\ 10 & 2 & 3 \\ \beta & 3 & 5 \end{array}\right| \\\\ & =\left|\begin{array}{ccc} 0 & 1 & 0 \\ -2 & 2 & 1 \\ \beta-18 & 3 & 2 \end{array}\right|=-1(-4-\beta+18) \\\\ & =\beta-14 \end{aligned} $$ <br/><br/>and <br/><br/>$\begin{aligned} \Delta_2 & =\left|\begin{array}{ccc}1 & 6 & 1 \\ 1 & 10 & 3 \\ 1 & \beta & 5\end{array}\right| \\\\ & =1(50-3 \beta)-6(5-3)+1(\beta-10) \\\\ & =50-3 \beta-12+\beta-10 \\\\ & =28-2 \beta=2(14-\beta)\end{aligned}$ <br/><br/> and <br/><br/>$$ \begin{aligned} \Delta_3 & =\left|\begin{array}{ccc} 1 & 1 & 6 \\ 1 & 2 & 10 \\ 1 & 3 & \beta \end{array}\right| \\ & =1(2 \beta-30)-1(\beta-10)+6(3-2) \\\\ & =2 \beta-30-\beta+10+6 \\\\ & =\beta-14 \end{aligned} $$ <br/><br/>Thus, at $\beta=14, \Delta_1=\Delta_2=\Delta_3=0$ <br/><br/>$\Rightarrow \alpha=3, \beta=14$ <br/><br/>So, system has infinite solutions.