If a point $P (\alpha, \beta, \gamma)$ satisfying $(\alpha\,\, \beta\,\, \gamma) \begin{pmatrix} 2 & 10 & 8 \\9 & 3 & 8 \\8 & 4 & 8\end{pmatrix}=(0\,\,0\,\,0) $ lies on the plane $2 x+4 y+3 z=5$, then $6 \alpha+9 \beta+7 \gamma$ is equal to :
The absolute difference of the coefficients of \(x^{10}\) and \(x^7\) in the expansion of \(\left(2x^2 + \frac{1}{2x}\right)^{11}\) is equal to:
The number of points on the curve \(y=54 x^5-135 x^4-70 x^3+180 x^2+210 x\) at which the normal lines are parallel \(to x+90 y+2=0\) is
Consider the lines $L_1$ and $L_2$ given by
$L_1: \frac{x-1}{2}=\frac{y-3}{1}=\frac{z-2}{2} $
$ L_2: \frac{x-2}{1}=\frac{y-2}{2}=\frac{z-3}{3}$
A line $L_3$ having direction ratios $1,-1,-2$, intersects $L_1$ and $L_2$ at the points $P$ and $Q$ respectively Then the length of line segment $P Q$ is
The sum of the absolute maximum and minimum values of the function \(f(x)=\left|x^2-5 x+6\right|-3 x+2\)in the interval \([-1,3]\) is equal to :
Shortest distance between lines \(\frac{(x-5)}{4}\)=\(\frac{(y-3)}{6}\)=\(\frac{(z-2)}{4}\) and \(\frac{(x-3)}{7}=\frac{(y-2)}{5}=\frac{(z-9)}{6}\) is ?
If (21)18 + 20·(21)17 + (20)2 · (21)16 + ……….. (20)18 = k (2119 – 2019) then k =
Let the system of linear equations$-x + 2y - 9z = 7$,$-x + 3y + 72 = 9$,$-2x + y + 5z = 8$,$-3x + y + 13z = \lambda$has a unique solution $x = \alpha, y = \beta, z = \gamma$. Then the distance of the point $(\alpha, \beta, \gamma)$ from the plane $2x - 2y + z = \lambda$ is:
Let $\vec{a}, \vec{b}, \vec{c}$ be three vectors such that $|\vec{a}|=\sqrt{31}, 4|\vec{b}|=|\vec{c}|=2$ and $2(\vec{a} \times \vec{b})=3(\vec{c} \times \vec{a})$ If the angle between $\vec{b}$ and $\vec{c}$ is $\frac{2 \pi}{3}$, then $\left(\frac{\vec{a} \times \vec{c}}{\vec{a} \cdot \vec{b}}\right)^2$ is equal to _____