Let $p$ and $p+2$ be prime numbers and let $\Delta=\left|\begin{array}{ccc}p ! & (p+1) ! & (p+2) ! \\ (p+1) ! & (p+2) ! & (p+3) ! \\ (p+2) ! & (p+3) ! & (p+4) !\end{array}\right|$ Then the sum of the maximum values of $\alpha$ and $\beta$, such that $p ^{\alpha}$ and $( p +2)^{\beta}$ divide $\Delta$, is $........$

  • [JEE MAIN 2022]
  • A

    $4$

  • B

    $3$

  • C

    $2$

  • D

    $1$

Similar Questions

The value of the determinant $\left| {\,\begin{array}{*{20}{c}}1&2&3\\3&5&7\\8&{14}&{20}\end{array}\,} \right|$is

Let $\alpha, \beta$ and $\gamma$ be real numbers. consider the following system of linear equations

$x+2 y+z=7$

$x+\alpha z=11$

$2 x-3 y+\beta z=\gamma$

Match each entry in List - $I$ to the correct entries in List-$II$

List - $I$ List - $II$
($P$) If $\beta=\frac{1}{2}(7 \alpha-3)$ and $\gamma=28$, then the system has ($1$) a unique solution
($Q$) If $\beta=\frac{1}{2}(7 \alpha-3)$ and $\gamma \neq 28$, then the system has ($2$) no solution

($R$) If $\beta \neq \frac{1}{2}(7 \alpha-3)$ where $\alpha=1$ and $\gamma \neq 28$,

then the system has

($3$) infinitely many solutions
($S$) If $\beta \neq \frac{1}{2}(7 \alpha-3)$ where $\alpha=1$ and $\gamma=28$, then the system has ($4$) $x=11, y=-2$ and $z=0$ as a solution
  ($5$) $x=-15, y=4$ and $z=0$ as a solution

Then the system has

  • [IIT 2023]

The set of all values of $\lambda$ for which the system of linear  $2{x_1} - 2{x_2} + {x_3} = \lambda {x_1}\;,\;2{x_1} - 3{x_2} + 2{x_3} = \lambda {x_2}\;\;,$$\;\; - {x_1} + 2{x_2} = \lambda {x_3}$ has a non-trivial solution

  • [JEE MAIN 2015]

Let $\omega = - \frac{1}{2} + i\frac{{\sqrt 3 }}{2}$. Then the value of the determinant $\left| {\,\begin{array}{*{20}{c}}1&1&1\\1&{ - 1 - {\omega ^2}}&{{\omega ^2}}\\1&{{\omega ^2}}&{{\omega ^4}}\end{array}\,} \right|$ is

  • [IIT 2002]

If $\alpha \neq \mathrm{a}, \beta \neq \mathrm{b}, \gamma \neq \mathrm{c}$ and $\left|\begin{array}{lll}\alpha & \mathrm{b} & \mathrm{c} \\ \mathrm{a} & \beta & \mathrm{c} \\ \mathrm{a} & \mathrm{b} & \gamma\end{array}\right|=0$,then $\frac{a}{\alpha-a}+\frac{b}{\beta-b}+\frac{\gamma}{\gamma-c}$ is equal to :

  • [JEE MAIN 2024]