If $\left| {\,\begin{array}{*{20}{c}}{x – 1}&3&0\\2&{x – 3}&4\\3&5&6\end{array}\,} \right| = 0$, then $x =$

Find area of the triangle with vertices at the point given in each of the following: $(2,7),(1,1),(10,8)$

If ${a_1},{a_2},{a_3}…..{a_n}….$ are in $G.P.$ then the value of the determinant $\left| {\,\begin{array}{*{20}{c}}{\log {a_n}}&{\log {a_{n + 1}}}&{\log {a_{n + 2}}}\\{\log {a_{n + 3}}}&{\log {a_{n + 4}}}&{\log {a_{n + 5}}}\\{\log {a_{n + 6}}}&{\log {a_{n + 7}}}&{\log {a_{n + 8}}}\end{array}\,} \right|$ is

Consider the system of linear equations ${a_1}x + {b_1}y + {c_1}z + {d_1} = 0$, ${a_2}x + {b_2}y + {c_2}z + {d_2} = 0$ and ${a_3}x + {b_3}y + {c_3}z + {d_3} = 0$. Let us denote by $\Delta (a,b,c)$ the determinant $\left| {\,\begin{array}{*{20}{c}}{{a_1}}&{{b_1}}&{{c_1}}\\{{a_2}}&{{b_2}}&{{c_2}}\\{{a_3}}&{{b_3}}&{{c_3}}\end{array}\,} \right|$ if $\Delta (a,b,c) \ne 0$, then the value of $x$ in the unique solution of the above equations is

If $\left| {\begin{array}{*{20}{c}}
{a – b – c}&{2a}&{2a}\\
{2b}&{b – c – a}&{2b}\\
{2c}&{2c}&{c – a – b}
\end{array}} \right|$ $ = \left( {a + b + c} \right)\,{\left( {x + a + b + c} \right)^2}$ , $x   \ne 0$ and $a + b + c \ne 0$, then $x$ is equal to