$\left| {\,\begin{array}{*{20}{c}}{a + b}&{a + 2b}&{a + 3b}\\{a + 2b}&{a + 3b}&{a + 4b}\\{a + 4b}&{a + 5b}&{a + 6b}\end{array}\,} \right| = $

$\left| {\,\begin{array}{*{20}{c}}1&{1 + ac}&{1 + bc}\\1&{1 + ad}&{1 + bd}\\1&{1 + ae}&{1 + be}\end{array}\,} \right| = $

Using properties of determinants, prove this:

$\left|\begin{array}{ccc}1 & 1+p & 1+p+q \\ 2 & 3+2 p & 4+3 p+2 q \\ 3 & 6+3 p & 10+6 p+3 q\end{array}\right|=1$

The number of distinct real roots of the equaiton, $\left|\begin{array}{*{20}{c}}
{\cos \,\,x}&{\sin \,\,x}&{\sin \,\,x}\\
{\sin \,\,x}&{\cos \,\,x}&{\sin \,\,x}\\
{\sin \,\,x}&{\sin \,\,x}&{\cos \,\,x}
\end{array}\right|\,\, = \,\,0$ in the interval $\left[ { – \frac{\pi }{4},\frac{\pi }{4}} \right]$ is

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 $…..$