If the three points $A(1; a; b)$ $B(a;2 ;b)$ $C(a; b; 3)$ are aligned, what is the value of $a+b$ ? I think that the range of this matrix should be equal to one: $$\begin{pmatrix} x_b-x_a & y_b-y_a & z_b-z_a\\ x_c-x_a& y_c-y_a & z_c-z_a \end{pmatrix} \quad$$ $$\begin{pmatrix} a-1 & 2-a & 0 \\ a-1 & b-a & 3-b \end{pmatrix} \quad$$ As a consequence the determinants of the submatrices $2\times2$ should be zero. $$\begin{pmatrix} a-1 & 0\\ a-1 & 3-b \end{pmatrix} \quad$$ $$\begin{pmatrix} 2-a & 0 \\ b-a & 3-b \end{pmatrix} \quad$$ $$\begin{pmatrix} a-1 & 2-a \\ a-1 & b-a \end{pmatrix} \quad$$ We have $$(a-1)(3-b)=0$$ $$(a-2)(3-b)=0$$ $$(a-1)(b-a)-(2-a)(a-1)=0$$ so $a=1$ and $b=3$
We get $a+b=4$ is that right ?
You have the formula for determinant wrong. It should be $(a-1)(3-b)=0$. Also, look at you last equation:$$2-a=b-3$$ From here you get $a+b=5$. This is wrong
You have three submatrices, not 2 $$\begin{pmatrix} a-1 & 0\\ a-1 & 3-b \end{pmatrix} $$ $$\begin{pmatrix} 2-a & 0 \\ b-a & 3-b \end{pmatrix} $$ $$\begin{pmatrix} a-1 &2-a \\ a-1&b-a \end{pmatrix} $$ Then from the determinant of the first matrix you get $a=1$ or $b=3$. In the second determinant you get $a=2$ or $b=3$. Obviously these two determinants can be $0$ only if $b=3$. In the last determinant you get $$(a-1)[(3-a)-(2-a)]=0$$ so $a=1$