$A$ point $P$ lies on a line passing through $Q(1, -2, 3)$ and is parallel to the line $\frac{x}{1} = \frac{y}{4} = \frac{z}{5}$. If $P$ lies on the plane $2x + 3y - 4z + 22 = 0$, then the length of the segment $PQ$ is:

  • A
    $\sqrt{42}$ units
  • B
    $\sqrt{32}$ units
  • C
    $4$ units
  • D
    $5$ units

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Similar Questions

If $P$,$Q$,and $R$ are the feet of the perpendiculars drawn from the point $A(1, 1, 1)$ to the planes $P_1: x + 2y + 2z = 2$,$P_2: 2x - 2y + z = -8$,and to the line of intersection of $P_1$ and $P_2$ respectively,then the area of $\Delta PQR$ is:

If the distance of the point $P(1, -2, 1)$ from the plane $x + 2y - 2z = \alpha$, where $\alpha > 0$, is $5$, then the foot of the perpendicular from $P$ to the plane is

The equation of the plane passing through the line of intersection of the planes $\overline{r} \cdot(2 \hat{i}-3 \hat{j}+4 \hat{k})=1$ and $\overline{r} \cdot(\hat{i}-\hat{j})+4=0$,and perpendicular to the plane $\overline{r} \cdot(2 \hat{i}-\hat{j}+\hat{k})+8=0$,is given by $\overline{r} \cdot(-5 \hat{i}+2 \hat{j}+12 \hat{k})=\mu$. Then the value of $\mu$ is:

Let the lines $\frac{x-1}{\lambda}=\frac{y-2}{1}=\frac{z-3}{2}$ and $\frac{x+26}{-2}=\frac{y+18}{3}=\frac{z+28}{\lambda}$ be coplanar and $P$ be the plane containing these two lines. Then which of the following points does $NOT$ lie on $P$?

The symmetric equation of the line formed by the intersection of the planes $3x + 2y + z - 5 = 0$ and $x + y - 2z - 3 = 0$ is:

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