Equation of Line: Symmetrical Form

Q. The equation of the line passing through $(-4,3,1)$, parallel to the plane $x+2y-z-5=0$ and intersecting the line $\frac{x+1}{-3}=\frac{y-3}{2}=\frac{z-2}{-1}$ is:

1. $\frac{x+4}{-1}=\frac{y-3}{1}=\frac{z-1}{1}$
2. $\frac{x+4}{1}=\frac{y-3}{1}=\frac{z-1}{3}$
3. $\frac{x+4}{3}=\frac{y-3}{-1}=\frac{z-1}{1}$
4. $\frac{x+4}{2}=\frac{y-3}{1}=\frac{z-1}{4}$

Q.

The distance of the point (1, 0, 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

1. $2\sqrt{14}$

2. 8

3. $3\sqrt{21}$

4. 13

Q.

If the lines $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$ and $\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}$ intersect, then the value of k is -

1. $\frac{3}{2}$

2. $\frac{9}{2}$

3. $-\frac{2}{9}$

4. $-\frac{3}{2}$

Q. The point on the line $\frac{x-2}{1}=\frac{y+3}{-2}=\frac{z+5}{-2}$ at a distance of 6 from the point (2, -3, -5) is

1. (3, -5, -3)
2. (4, -7, -9)
3. (0, 2, -1)
4. (-3, 5, 3)

Q. If the lines $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$ and $\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}$ intersect, then k =

1. $\frac{2}{9}$
2. $\frac{9}{2}$
3. \N
4. 3

Q. The equation of line $AB$ is $\frac{x}{2}=\frac{y}{-3}=\frac{z}{6}$. Through a point $P(1,2,5)$, line $PN$ is drawn perpendicular to $AB$ and line $PQ$ is drawn parallel to the plane $3x+4x+5z=0$ to meet $AB$ at $Q$. Then

1. Coordinates of $N$ are $(\frac{52}{49},-\frac{78}{49},\frac{156}{49})$
2. Equation of line $NQ$ is $3(x-3)=-(2y+9)=z-9$
3. Equation of line $NQ$ is $3(x-3)=(2y+9)=z-9$
4. Coordinates of $Q$ are $(3,\frac{9}{2},9)$