Let the system of linear equations

$x+y+\alpha z=2$

$3 x+y+z=4$

$x+2 z=1$

have a unique solution $\left(x^{*}, y^{*}, z^{*}\right)$. If $\left(\alpha, x^{*}\right),\left(y^{*}, \alpha\right)$ and $\left(x^{*},-y^{*}\right)$ are collinear points, then the sum of absolute values of all possible values of $\alpha$ is

Find values of $x$, if $\left|\begin{array}{ll}2 & 4 \\ 5 & 1\end{array}\right|=\left|\begin{array}{cc}2 x & 4 \\ 6 & x\end{array}\right|$

The value of $\left| {\,\begin{array}{*{20}{c}}{{1^2}}&{{2^2}}&{{3^2}}\\{{2^2}}&{{3^2}}&{{4^2}}\\{{3^2}}&{{4^2}}&{{5^2}}\end{array}\,} \right|$ is

Which of the following is correct?

For non zero, $a,b,c$ if $\Delta = \left| {\,\begin{array}{*{20}{c}}{1 + a}&1&1\\1&{1 + b}&1\\1&1&{1 + c}\end{array}} \right| = 0$, then the value of $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = $

Consider system of equations in $x$ , $y$ and $z$

$12x + by + cz = 0$ ;   $ax + 24y + cz = 0$  ;   $ax + by + 36z = 0$ .

(where $a$ , $b$ , $c$ are real numbers, $a \ne 12$ , $b \ne 24$ , $c \ne 36$ ).

If system of equation has solution and $z \ne 0$, then value of  $\frac{1}{{a - 12}} + \frac{2}{{b - 24}} + \frac{3}{{c - 36}}$ is