Let $a, b, c, d$ be in arithmetic progression with common difference $\lambda$. If

$\left|\begin{array}{lll} x+a-c & x+b & x+a \\ x-1 & x+c & x+b \\ x-b+d & x+d & x+c \end{array}\right|=2$

then value of $\lambda^{2}$ is equal to $.....$

If $\left| {\,\begin{array}{*{20}{c}}{x + 1}&{x + 2}&{x + 3}\\{x + 2}&{x + 3}&{x + 4}\\{x + a}&{x + b}&{x + c}\end{array}\,} \right| = 0$, then $a,b,c$ are in

The value of $\left| {\,\begin{array}{*{20}{c}}{441}&{442}&{443}\\{445}&{446}&{447}\\{449}&{450}&{451}\end{array}\,} \right|$ is

Using properties of determinants, prove that:

$\left|\begin{array}{ccc}\alpha & \alpha^{2} & \beta+\gamma \\ \beta & \beta^{2} & \gamma+\alpha \\ \gamma & \gamma^{2} & \alpha+\beta\end{array}\right|=(\beta-\gamma)(\gamma-\alpha)(\alpha-\beta)(\alpha+\beta+\gamma)$

Let the numbers $2, b, c$ be in an $A.P$ and $A = \left[ {\begin{array}{*{20}{c}}
  1&1&1 \\ 
  2&b&c \\ 
  4&{{b^2}}&{{c^2}} 
\end{array}} \right]$. If $det(A) \in [2,16]$ then $c$ lies in the interval

If $a^2 + b^2 + c^2 = - 2$ and $f (x) = $ $\left| {\,\begin{array}{*{20}{c}}{1 + {a^2}x}&{(1 + {b^2})x}&{(1 + {c^2})x}\\{(1 + {a^2})x}&{1 + {b^2}x}&{(1 + {c^2})x}\\ {(1 + {a^2})x}&{(1 + {b^2})x}&{1 + {c^2}x}\end{array}\,} \right|$ then $f (x)$ is a polynomial of degree