The system of equations $\lambda x + y + z = 0, -x + \lambda y + z = 0, -x - y + \lambda z = 0$ will have a non-zero solution if real values of $\lambda$ are given by

If $x$ is a positive integer,then $\Delta = \left| \begin{array}{ccc} x! & (x+1)! & (x+2)! \\ (x+1)! & (x+2)! & (x+3)! \\ (x+2)! & (x+3)! & (x+4)! \end{array} \right|$ is equal to

Evaluate the determinant: $\left| {\begin{array}{*{20}{c}}{b + c}& a& a\\b& {c + a}& b\\c& c& {a + b}\end{array}} \right|$

Show that points $A(a, b+c), B(b, c+a), C(c, a+b)$ are collinear.

Let $f(\theta) = \left| \begin{array}{ccc} 1 & \cos \theta & -1 \\ -\sin \theta & 1 & -\cos \theta \\ -1 & \sin \theta & 1 \end{array} \right|$. Suppose $A$ and $B$ are respectively the maximum and minimum values of $f(\theta)$. Then $(A, B)$ is equal to

If $\left|\begin{array}{ccc}2 a & x_{1} & y_{1} \\ 2 b & x_{2} & y_{2} \\ 2 c & x_{3} & y_{3}\end{array}\right|=\frac{a b c}{2} \neq 0$,then the area of the triangle whose vertices are $\left(\frac{x_{1}}{a}, \frac{y_{1}}{a}\right), \left(\frac{x_{2}}{b}, \frac{y_{2}}{b}\right), \left(\frac{x_{3}}{c}, \frac{y_{3}}{c}\right)$ is: