Let the image of the point $P(2, -1, 3)$ in the plane $x + 2y - z = 0$ be $Q$. Then the distance of the plane $3x + 2y + z + 29 = 0$ from the point $Q$ is $.........$.

  • A
    $\frac{22 \sqrt{2}}{7}$
  • B
    $\frac{24 \sqrt{2}}{7}$
  • C
    $2 \sqrt{14}$
  • D
    $3 \sqrt{14}$

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The distance of the point $(1, 6, 2)$ from the point of intersection of the line $\frac{x-2}{3} = \frac{y+1}{4} = \frac{z-2}{12}$ and the plane $x-y+z=16$ is (in $\text{ units}$)

Find the equation of the plane passing through the points $(2, 1, -1)$ and $(-1, 3, 4)$ and perpendicular to the plane $x - 2y + 4z = 10$.

If $\vec{a}=\hat{i}-\hat{j}+3 \hat{k}$ and $\vec{c}=-\hat{k}$ are position vectors of two points, and $\vec{b}=2 \hat{i}-\hat{j}+\lambda \hat{k}$ and $\vec{d}=\hat{i}+2 \hat{j}-\hat{k}$ are two vectors, then the lines $\vec{r}=\vec{a}+t \vec{b}$ and $\vec{r}=\vec{c}+s \vec{d}$ are:

In $R^3$,consider the planes $P_1: y=0$ and $P_2: x+z=1$. Let $P_3$ be a plane,different from $P_1$ and $P_2$,which passes through the intersection of $P_1$ and $P_2$. If the distance of the point $(0,1,0)$ from $P_3$ is $1$ and the distance of a point $(\alpha, \beta, \gamma)$ from $P_3$ is $2$,then which of the following relations is (are) true?
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