An equilateral triangle has each of its sides of length $6 \text{ cm}$. If $(x_1, y_1), (x_2, y_2), \text{ and } (x_3, y_3)$ are its vertices,then the value of the determinant $\left| \begin{array}{ccc} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{array} \right|^2$ is equal to:
- A$192$
- B$243$
- C$486$
- D$972$
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Similar Questions
If $ax^4+bx^3+cx^2+50x+d = \begin{vmatrix} x^3-14x^2 & -x & 3x+\lambda \\ 4x+1 & 3x & x-4 \\ -3 & 4 & 0 \end{vmatrix}$,then find $\lambda$.
DifficultAP EAMCET 2020
View Solution$\left|\begin{array}{ccc}\cos 3\pi & \sin 5\pi & \tan 7\pi \\ \sqrt{3} & 1 & 0 \\ \sqrt{5} & 0 & 1\end{array}\right| = $ . . . . . . .
Given that,$a \alpha^2+2 b \alpha+c \neq 0$ and that the system of equations
$\begin{aligned} & (a \alpha+b) x+a y+b z=0 \\ & (b \alpha+c) x+b y+c z=0 \\ & (a \alpha+b) y+(b \alpha+c) z=0\end{aligned}$
has a non-trivial solution,then $a, b$ and $c$ lie in
$\begin{aligned} & (a \alpha+b) x+a y+b z=0 \\ & (b \alpha+c) x+b y+c z=0 \\ & (a \alpha+b) y+(b \alpha+c) z=0\end{aligned}$
has a non-trivial solution,then $a, b$ and $c$ lie in
If $a_i^2 + b_i^2 + c_i^2 = 1$ for $i = 1, 2, 3$ and $a_ia_j + b_ib_j + c_ic_j = 0$ for $i \ne j$ where $i, j = 1, 2, 3$,then the value of the determinant $\left| \begin{array}{ccc} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{array} \right|$ is:
Difficult
View SolutionThe value of $\left| \begin{array}{ccc} 1 & 1 & 1 \\ bc & ca & ab \\ b+c & c+a & a+b \end{array} \right|$ is
Medium
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