Foot of Perpendicular of a Point on a Line

Q. The distance of the point $P(-2,3,-4)$ from the line $\frac{x+2}{3}=\frac{2y+3}{4}=\frac{3z+4}{5}$ measured parallel to the plane $4x+12y-3z+1=0$ is $d$, then the value of $(2d-8)$ is

Q.

If A, B, C are points (1, 0, 4) (0, -1, 5) and (2, -3, 1) respectively, then the coordinates of the foot of the perpendicular drawn from A to the line BC are

1. $(\frac{1}{2},-\frac{1}{2},\frac{9}{2})$

2. $(1,-2,3)$

3. $(\frac{3}{2},-\frac{3}{2},\frac{5}{2})$

4. None of these

Q. If $z_1$ and $z_2$ lie on the same side the line $\overline{a}z+z\overline{z}+b=0$, where $a\in C,a\ne 0,b\in R$, then the ratio $\frac{\overline{a}{z}_1+a\overline{z_1}+b}{\overline{a}{z}_2+a\overline{z_2}+b}$ is

1. Purely imaginary
2. A negative real number
3. A positive real number
4. None of these

Q. The foot of the perpendicular from P(1, 0, 2) to the line $\frac{x+1}{3}=\frac{y-2}{-2}=\frac{z+1}{-1}$ is

1. $(1,2,-3)$
2. $(\frac{1}{2},1,-\frac{3}{2})$
3. $(2,4,-6)$
4. $(2,3,6)$