### Three Dimensional Geometry Question 17

#### Question 17 - 29 January - Shift 2

If the lines $\frac{x-1}{1}=\frac{y-2}{2}=\frac{z+3}{1}$ and $\frac{x-a}{2}=\frac{y+2}{3}=\frac{z-3}{1}$ intersects at the point $P$, then the distance of the point $P$ from the plane $z=a$ is :

(1) 16

(2) 28

(3) 10

(4) 22

## Show Answer

#### Answer: (2)

#### Solution:

#### Formula: A plane and a point

$ \begin{aligned} & \text { Point on } L_1 \equiv(\lambda+1,2 \lambda+2, \lambda-3) \\ & \text { Point on } L_2 \equiv(2 \mu+a, 3 \mu-2, \mu+3) \\ & \lambda-3=\mu+3 \quad \Rightarrow \lambda=\mu+6 \quad \ldots (1)\\ & 2 \lambda+2=3 \mu-2 \Rightarrow 2 \lambda=3 \mu-4 \quad \ldots (2) \end{aligned} $

Solving, (1) and (2)

$ \begin{array}{ll} \Rightarrow & \lambda=22 ; ;\text{and} ; ;\mu=16 \\ \Rightarrow & P \equiv(23,46,19) \\ \Rightarrow & a=-9 \end{array} $

Distance of P from $z=-9$ is 28