Plane and a Point

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

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

The distance of the point $\left(-1,2,3\right)$ from the plane $\overrightarrow{r}·(\hat{i}-2\hat{j}+3\hat{k})=10$ parallel to the line of the shortest distance between the lines$\overrightarrow{r}=(\hat{i}-\hat{j})+\lambda (2\hat{i}+\hat{k})$ and $\overrightarrow{r}=(2\hat{i}-\hat{j})+\mu (\hat{i}-\hat{j}+\hat{k})$ is

Let $(\alpha ,\beta ,\gamma )$ be the image of point $P(2,3,5)$ in the plane $2x+y-3z=6$. Then $\alpha +\beta +\gamma$ is equal to

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 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

The plane, passing through the points $(0, –1, 2)$ and $(–1, 2, 1)$ and parallel to the line passing through $(5, 1, –7)$ and $(1, –1, –1),$also passes through the point

Let the image of the point $\mathrm{P}(1, 2, 6)$in the plane passing through the points $\mathrm{A}(1, 2, 0)$ and $\mathrm{B}(1, 4, 1) \mathrm{C}(0, 5, 1)$ be $\mathrm{Q}(\mathrm{\alpha },\mathrm{\beta },\mathrm{\gamma })$ . Then $\left({\mathrm{\alpha }}^2+{\mathrm{\beta }}^2+{\mathrm{\gamma }}^2\right)$ equal to

Let two vertices of a triangle $ABC$ be $\left(2, 4, 6\right)$ and $\left(0,-2,-5\right)$, and its centroid be $\left(2,1,-1\right)$. If the image of the third vertex in the plane $x+2y+4z=11$ is $\left(\alpha , \beta , \gamma \right)$, then $\alpha \beta +\beta \gamma +\gamma \alpha$ is equal to

The distance of the point $\left(1,-2,3\right)$ from the plane $x-y+z=5$ measured parallel to the line, $\frac{x}{2}=\frac{y}{3}=\frac{z}{-6}$ is:

The distance of the point whose position vector is $\left(2\hat{i}+\hat{j}-\hat{k}\right)$ from the plane $\overrightarrow{r}·\left(\hat{i}-2\hat{j}+4\hat{k}\right)=4$ is

Let the foot of the perpendicular drawn from the point $\left(1,2,3\right)$ to a plane be $\left(-1,3,-2\right)$. Then the perpendicular distance from the origin to the plane is

The $x$-intercept of a plane $\pi$ passing through the point $\left(1,1,1\right)$ is $\frac{5}{2}$ and the perpendicular distance from the origin to the plane $\pi$ is $\frac{5}{7}$. If the $y$-intercept of the plane $\pi$ is negative and the $z$-intercept is positive then its $y$-intercept is

If the equation of the plane which is at a distance of $\frac{1}{3}$ units from the origin and perpendicular to a line whose directional ratios are $\left(1,2,2\right)$ is $x+py+qz+r=0$ then $\sqrt{p^2+q^2+r^2}=$

The distance from the point $\left(2,2,2\right)$ to the plane $2x-y+3z=5$ is equal to

Let $P_1: 3y+z+1=0$ and $P_2: 2x-y+3z-7=0$ and the equation of line $AB$ is $\frac{x-1}{2}=\frac{y-3}{-1}=\frac{z-4}{3}$ in $3D$ space. Shortest distance between the line of intersection of plane $P_1$ and $P_2$ and the line $AB$ is equal to