\(a, b, c\) are distinct real numbers, not equal to one. If \(ax + y + z = 0\), \(x + by + z = 0\), and \(x + y + cz = 0\) have a non-trivial solution, then the value of \(\dfrac{1}{1-a} + \dfrac{1}{1-b} + \dfrac{1}{1-c}\) is equal to
Let a1, a2, a3, ..., a10 be in G.P. with ai > 0 for i = 1,2,..., 10 and S be the set of pairs (r, k), r, k ∈ N (the set of natural numbers) for which . Then the number of elements in S, is :
Let $p$ be an odd prime number and $T_p$ be the following set of $2 \times 2$ matrices: $T_p = \left\{ A = \begin{bmatrix} a & b \\ c & a \end{bmatrix} : a, b, c \in \{0, 1, 2, \dots, p-1\} \right\}$. The number of $A$ in $T_p$ such that $A$ is either symmetric or skew-symmetric or both, and $\det(A)$ divisible by $p$ is -