Let $D_1 = \begin{vmatrix} a & b & a+b \\ c & d & c+d \\ a & b & a-b \end{vmatrix}$ and $D_2 = \begin{vmatrix} a & c & a+c \\ b & d & b+d \\ a & c & a+b+c \end{vmatrix}$. Then the value of $\frac{D_1}{D_2}$,where $b \neq 0$ and $ad \neq bc$,is:

Evaluate $\left|\begin{array}{ccc}102 & 18 & 36 \\ 1 & 3 & 4 \\ 17 & 3 & 6\end{array}\right|$

Let $a, b, c$ be such that $b + c \ne 0$. If $\left| \begin{array}{ccc} a & a+1 & a-1 \\ -b & b+1 & b-1 \\ c & c-1 & c+1 \end{array} \right| + \left| \begin{array}{ccc} a+1 & b+1 & c-1 \\ a-1 & b-1 & c+1 \\ (-1)^{n+2} \cdot a & (-1)^{n+1} \cdot b & (-1)^n \cdot c \end{array} \right| = 0$,then $n$ is equal to:

$\left| \begin{array}{ccc} 11 & 12 & 13 \\ 12 & 13 & 14 \\ 13 & 14 & 15 \end{array} \right| = $

Using the property of determinants and without expanding,prove that:
$\left|\begin{array}{lll}b+c & q+r & y+z \\ c+a & r+p & z+x \\ a+b & p+q & x+y\end{array}\right|=2\left|\begin{array}{lll}a & p & x \\ b & q & y \\ c & r & z\end{array}\right|$

If $a \neq p, b \neq q, c \neq r$ and $\left|\begin{array}{ccc}p & b & c \\ p+a & q+b & 2c \\ a & b & r\end{array}\right|=0$, then $\frac{p}{p-a}+\frac{q}{q-b}+\frac{r}{r-c}$ is equal to :