The solution set of $\left|\begin{array}{ccc}x & 3 & 5 \\ 2 & 6 & 10 \\ 7 & 21 & 35\end{array}\right|=0$ is . . . . . . .

Let $A = \begin{bmatrix} [x+1] & [x+2] & [x+3] \\ [x] & [x+3] & [x+3] \\ [x] & [x+2] & [x+4] \end{bmatrix}$,where $[t]$ denotes the greatest integer less than or equal to $t$. If $\operatorname{det}(A) = 192$,then the set of values of $x$ is the interval:

If the inverse of the matrix $\begin{bmatrix} \alpha & 14 & -1 \\ 2 & 3 & 1 \\ 6 & 2 & 3 \end{bmatrix}$ does not exist,then the value of $\alpha$ is:

The values of $\alpha$,for which $\left|\begin{array}{ccc}1 & \frac{3}{2} & \alpha+\frac{3}{2} \\ 1 & \frac{1}{3} & \alpha+\frac{1}{3} \\ 2 \alpha+3 & 3 \alpha+1 & 0\end{array}\right|=0$,lie in the interval

If $A = \begin{bmatrix} 1 & 0 & 1 \\ 2 & 1 & 0 \\ 3 & 2 & 1 \end{bmatrix}$,then $\det A$ =

Find the values of $x$ for which $\left|\begin{array}{ll}3 & x \\ x & 1\end{array}\right|=\left|\begin{array}{ll}3 & 2 \\ 4 & 1\end{array}\right|$.