Using the property of determinants and without expanding, Prove that 

$\left|\begin{array}{lll}x & a & x+a \\ y & b & y+b \\ z & c & z+c\end{array}\right|=0$

Let $M$ be a $3 \times 3$ invertible matrix with real entries and let $I$ denote the $3 \times 3$ identity matrix. If $M ^{-1}=\operatorname{adj}(\operatorname{adj} M )$, then which of the following statement is/are $ALWAYS TRUE$ ?

$(A)$ $M=I$   $(B)$ $\operatorname{det} M =1$   $(C)$ $M ^2= I$  $(D)$ $(\operatorname{adj} M)^2=I$

If $\left| {\begin{array}{*{20}{c}}   {a - b}&{b - c}&{c - a} \\    {b - c}&{c - a}&{a - b} \\    {c - a + 1}&{a - b}&{b - c}  \end{array}} \right| = 0$ ,$\left( {a,b,c \in R - \left\{ 0 \right\}} \right),$ then

If $\left| {\,\begin{array}{*{20}{c}}{y + z}&x&y\\{z + x}&z&x\\{x + y}&y&z\end{array}\,} \right| = k(x + y + z){(x - z)^2}$, then $k = $

If $a, b, c$ are real then the value of determinant $\left| {\begin{array}{*{20}{c}} {{a^2} + 1}&{ab}&{ac}\\{ab}&{{b^2} + 1}&{bc}\\{ac}&{bc}&{{c^2} + 1}\end{array}}\right|$ $= 1$ if

Using the property of determinants and without expanding, prove that:

$\left|\begin{array}{lll}1 & b c & a(b+c) \\ 1 & c a & b(c+a) \\ 1 & a b & c(a+b)\end{array}\right|=0$