Three-Dimensional Geometry

The plane containing the line $\dfrac{x-1}1=\dfrac{y-2}2=\dfrac{z-3}3$ and parallel to the line $\dfrac x1=\dfrac y1=\dfrac z4$ passes through the point :

A symmetrical form of the line of intersection of the planes $x=ay+b$ and $z=cy+d$ is :

If the distance between planes, $4x-2y-4z+1=0$ and $4x-2y-4z+d=0$ is 7, then d is :

Equation of the line of the shortest distance between the lines $\dfrac x1=\dfrac y{-1}=\dfrac z1$ and $\dfrac{x-1}0=\dfrac{y+1}{-2}=\dfrac z1$ is :

If the angle between the line $2(x+1)=y=z+4$ and the plane $2x-y+\sqrt\lambda z+4=0$ is $\dfrac\pi6$, then the value of λ is :

Perpendiculars are drawn from points on the line $\dfrac{x+2}2=\dfrac{y+1}{-1}=\dfrac z3$ to the plane $x+y+z=3$. The feet of perpendiculars lie on the line

A line $l$ passing through the origin is perpendicular to the lines
$$l_1:(3+t)\hat i+(-1+2t)\hat j+(4+2t)\hat k,\quad-\infty<t<\infty$$
$$l_2:(3+2s)\hat i+(3+2s)\hat j+(2+s)\hat k,\quad-\infty<s<\infty$$
Then, the coordinate(s) of the point(s) on $l_2$ at a distance of $\sqrt{17}$ from the point of intersection of $l$ and $l_1$ is (are)

Two lines $L_1:x=5,\ \dfrac y{3-\alpha}=\dfrac z{-2}$ and $L_2:x=\alpha,\ \dfrac y{-1}=\dfrac z{2-\alpha}$ are coplanar. Then $\alpha$ can take value(s)

Consider the lines $L_1:\dfrac{x-1}2=\dfrac y{-1}=\dfrac{z+3}1$, $L_2:\dfrac{x-4}1=\dfrac{y+3}1=\dfrac{z+3}2$ and the planes $P_1:7x+y+2z=3$, $P_2:3x+5y-6z=4$. Let $ax+by+cz=d$ be the equation of the plane passing through the point of intersection of lines $L_1$ and $L_2$, and perpendicular to planes $P_1$ and $P_2$.

Match List I with List II and select the correct answer using the code given below the lists:

The point $P$ is the intersection of the straight line joining the points $Q(2,3,5)$ and $R(1,-1,4)$ with the plane $5x-4y-z=1$. If $S$ is the foot of the perpendicular drawn from the point $T(2,1,4)$ to $QR$, then the length of the line segment $PS$ is

The equation of a plane passing through the line of intersection of the planes $x+2y+3z=2$ and $x-y+z=3$ and at a distance $\dfrac2{\sqrt3}$ from the point $(3,1,-1)$ is

If the straight lines $\dfrac{x-1}2=\dfrac{y+1}k=\dfrac z2$ and $\dfrac{x+1}5=\dfrac{y+1}2=\dfrac zk$ are coplanar, then the plane(s) containing these two lines is(are)

Equation of the plane containing the straight line $\dfrac x2=\dfrac y3=\dfrac z4$ and perpendicular to the plane containing the straight lines $\dfrac x3=\dfrac y4=\dfrac z2$ and $\dfrac x4=\dfrac y2=\dfrac z3$ is

If the distance between the plane $Ax-2y+z=d$ and the plane containing the lines $\dfrac{x-1}2=\dfrac{y-2}3=\dfrac{z-3}4$ and $\dfrac{x-2}3=\dfrac{y-3}4=\dfrac{z-4}5$ is $\sqrt6$, then $|d|$ 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

Match the statements in Column-I with the values in Column-II.

Let $P(3,2,6)$ be a point in space and $Q$ be a point on the line $$\vec r=(\hat i-\hat j+2\hat k)+\mu(-3\hat i+\hat j+5\hat k).$$ Then the value of $\mu$ for which the vector $\overrightarrow{PQ}$ is parallel to the plane $x-4y+3z=1$ is

A line with positive direction cosines passes through the point $P(2,-1,2)$ and makes equal angles with the coordinate axes. The line meets the plane
$$2x+y+z=9$$
at point $Q$. The length of the line segment $PQ$ equals

Consider three planes
$$P_1:x-y+z=1,\qquad P_2:x+y-z=-1,\qquad P_3:x-3y+3z=2.$$
Let $L_1,L_2,L_3$ be the lines of intersection of the planes $P_2$ and $P_3$, $P_3$ and $P_1$, and $P_1$ and $P_2$, respectively.

STATEMENT-1: At least two of the lines $L_1$, $L_2$ and $L_3$ are non-parallel.

and

STATEMENT-2: The three planes do not have a common point.

The unit vector perpendicular to both $L_1$ and $L_2$ is