Three digit numbers $x17, 3y6$ and $12z$ where $x, y, z$ are integers from $0$ to $9$, are divisible by a fixed constant $k$. Then the determinant $\left| {\,\begin{array}{*{20}{c}}x&3&1\\7&6&z\\1&y&2\end{array}\,} \right|$ + $48$ must be divisible by

Statement $-1 :$Determinant of a skew-symmetric matrix of order $3$ is zero

Statement $-2 :$ For any matrix $A,$ $\det \left( {{A^T}} \right) = {\rm{det}}\left( A \right)$ and $\det \left( { – A} \right) = – {\rm{det}}\left( A \right)$ Where $\det \left( A \right) = A$. Then :

If ${\left| {\,\begin{array}{*{20}{c}}4&1\\2&1\end{array}\,} \right|^2} = \left| {\,\begin{array}{*{20}{c}}3&2\\1&x\end{array}\,} \right| – \left| {\,\begin{array}{*{20}{c}}x&3\\{ – 2}&1\end{array}\,} \right|$, then $ x =$

$2x + 3y + 4z = 9$,$4x + 9y + 3z = 10,$$5x + 10y + 5z = 11$ then the value of $ x$ is

Two fair dice are thrown. The numbers on them are taken as $\lambda$ and $\mu$, and a system of linear equations

$x+y+z=5$    ;    $x+2 y+3 z=\mu$   ;     $x+3 y+\lambda z=1$

is constructed. If $\mathrm{p}$ is the probability that the system has a unique solution and $\mathrm{q}$ is the probability that the system has no solution, then :