$\left| \begin{array}{ccc} 0 & p-q & p-r \\ q-p & 0 & q-r \\ r-p & r-q & 0 \end{array} \right| = $

  • A
    $0$
  • B
    $(p-q)(q-r)(r-p)$
  • C
    $pqr$
  • D
    $3pqr$

Explore More

Similar Questions

The value of $\left|\begin{array}{ccc}1 & \log _{x} y & \log _{x} z \\ \log _{y} x & 1 & \log _{y} z \\ \log _{z} x & \log _{z} y & 1\end{array}\right|$ is equal to

Three-digit numbers $x17$,$3y6$,and $12z$,where $x, y, z$ are integers from $0$ to $9$,are divisible by a fixed constant $k$. Then the determinant $\left| \begin{array}{ccc} x & 3 & 1 \\ 7 & 6 & z \\ 1 & y & 2 \end{array} \right| + 48$ must be divisible by:

If $\left| \begin{array}{ccc} x - 1 & 3 & 0 \\ 2 & x - 3 & 4 \\ 3 & 5 & 6 \end{array} \right| = 0$,then $x =$

$\begin{aligned} & \text { If }\left|\begin{array}{ccc}n^2 & (n+1)^2 & (n+2)^2 \\ (n+1)^2 & (n+2)^2 & (n+3)^2 \\ (n+2)^2 & (n+3)^2 & (n+4)^2\end{array}\right|=\Delta \text { and } \\ & \left|\begin{array}{ccc}1 & -4 & 7 \\ -2 & 3 & -5 \\ 3 & x & -3\end{array}\right|=2 \Delta+1, \text { then } x=\end{aligned}$

If $\left| \begin{array}{ccc} x+1 & 3 & 5 \\ 2 & x+2 & 5 \\ 2 & 3 & x+4 \end{array} \right| = 0$,then $x =$

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo