Equation of a Plane and Related Concepts

Let the line $L:x=\frac{1-y}{-2}=\frac{z-3}{\lambda },\lambda \in \mathrm{ℝ}$ meet the plane $P: x+2 y+3 z=4$ at the point $(\alpha ,\beta ,\gamma )$. If the angle between the line $L$ and the plane $P$ is ${\cos }^{-1}\left(\sqrt{\frac{5}{14}}\right)$, then $\alpha +2\beta +6\gamma$ is equal to

Let $N$ be the foot of perpendicular from the point$P(1,-2,3)$ on the line passing through the points $(4,5,8)$ and $(1,-7,5)$. Then the distance of $N$ from the plane $2x-2y+z+5=0$ is

The line, that is coplanar to the line $\frac{x+3}{-3}=\frac{y-1}{1}=\frac{z-5}{5}$, is

Let the line passing through the points $P\left(2,-1,2\right)$ and $Q\left(5,3,4\right)$ meet the plane $x-y+z=4$ at the point $R$. Then the distance of the point $R$ from the plane $x+2y+3z+2=0$ measured parallel to the line $\frac{x-7}{2}=\frac{y+3}{2}=\frac{z-2}{1}$ is

Let $P$ be the plane passing through the points $\left(5,3,0\right),\left(13,3,-2\right)$ and $\left(1,6,2\right)$. For $\alpha \in \mathrm{\mathbb{N}}$, if the distance of the points$A\left(3,4,\alpha \right)$ and $B\left(2,\alpha ,a\right)$ from the plane $P$ are $2$ and $3$ respectively, then the positive value of $a$ is

Let the line $\frac{x}{1}=\frac{6-y}{2}=\frac{z+8}{5}$ intersect the lines $\frac{x-5}{4}=\frac{y-7}{3}=\frac{z+2}{1}$  and $\frac{x+3}{6}=\frac{3-y}{3}=\frac{z-6}{1}$ at the points $\mathrm{A}$ and $\mathrm{B}$ respectively. Then the distance of the mid-point of the line segment $AB$ from the plane $2x-2y+z=14$ is

Let the plane $P$ contain the line $2x+y-z-3=0=5x-3y+4z+9$ and be parallel to the line $\frac{x+2}{2}=\frac{3-y}{-4}=\frac{z-7}{5}$. Then the distance of the point $A\left(8, -1, -19\right)$ from the plane $P$ measured parallel to the line $\frac{x}{-3}=\frac{y-5}{4}=\frac{2-z}{-12}$ is equal to _________.

Let the foot of perpendicular from the point $A\left(4,3,1\right)$ on the plane $P: x-y+2z+3=0$ be $N$. If $B(5,\alpha ,\beta ),\alpha ,\beta \in \mathbb{Z}$ is a point on plane $P$ such that the area of the triangle $ABN$ is $3\sqrt{2}$, then ${\alpha }^2+{\beta }^2+\alpha \beta$ is equal to _____________

If the equation of the plane that contains the point $(-2,3,5)$ and is perpendicular to each of the planes $2x+4y+5z=8$ and $3x-2y+3z=5$ is $\alpha x+\beta y+\gamma z+97=0$ then $\alpha +\beta +\gamma =$

Let the system of linear equations

$–x+2y-9z=7$

$-x+3y+7z=9$

$-2x+y+5z=8$

$-3x+y+13z=\lambda$

has a unique solution $x=\alpha , y=\beta , z=\gamma$. Then the distance of the point $\left(\alpha , \beta , \gamma \right)$ from the plane $2x-2y+z=\lambda$ is

Let the equation of plane passing through the line of intersection of the planes $x+2y+az=2$ and $x-y+z=3$ be $5x-11y+bz=6a-1.$ For $c\in \mathrm{\mathbb{Z}},$ if the distance of this plane from the point $\left(a,-c,c\right)$ is $\frac{2}{\sqrt{a}},$ then $\frac{a+b}{c}$ is equal to

Let the foot of perpendicular of the point $P\left(3, –2, –9\right)$ on the plane passing through the points $\left(–1, –2, –3\right), \left(9, 3, 4\right), \left(9, –2, 1\right)$ be $Q\left(\alpha , \beta , \gamma \right)$. Then the distance $Q$ from the origin is

Let ${\lambda }_1, {\lambda }_2$ be the values of $\lambda$ for which the points $\left(\frac{5}{2},1, \lambda \right)$ and $\left(-2, 0, 1\right)$ are at equal distance  from the plane $2x+3y-6z+7$ If ${\lambda }_1>{\lambda }_2$

then the distance of the point $\left({\lambda }_1-{\lambda }_2,{\lambda }_2,{\lambda }_1\right)$ from the line $\frac{x-5}{1}=\frac{y-1}{2}=\frac{z+7}{2}$ is $\_\_\_\_\_\_$

For $a,b\in Z$ and $|a-b|\le 10$, let the angle between the plane $P: a x+y-z=b$ and the line $L: x-1=a-y=z+1$ be ${\cos }^{-1}\left(\frac{1}{3}\right)$ If the distance of the point $(6,-6,4)$ from the plane $P$ is $3\sqrt{6}$, then $a^4+b^2$ is equal to

Given $\frac{x+3}{-3}=\frac{y-2}{2}=\frac{z-5}{5}$. Which of the following lines in options is coplanar with the given line?