If $$a, b, c$$ are respectively the 5 th, 8 th, 13 th terms of an arithmetic progression, then $$\left|\begin{array}{ccc}a & 5 & 1 \\ b & 8 & 1 \\ c & 13 & 1\end{array}\right|=$$

If $$A=\left[\begin{array}{ccc}1 & 0 & 0 \\ a & -1 & 0 \\ b & c & 1\end{array}\right]$$ is such that $$A^2=I$$, then

Let $$A=\left[\begin{array}{ccc}-2 & x & 1 \\ x & 1 & 1 \\ 2 & 3 & -1\end{array}\right]$$. If the roots of the equation $$\operatorname{det} A=0$$ are $$l, m$$ then $$l^3-m^3=$$

For $$i=1,2,3$$ and $$j=1,23$$ If $$a_i^2+b_i^2+c_i^2=1, a_i a_j+b_i b_j+c_i c_j=0, \forall i \neq j$$ and $$A=\left[\begin{array}{lll}a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3\end{array}\right]$$, then $$\operatorname{det}\left(A A^T\right)=$$