If the image of a point \(P(3,2,1)\) in the line \[ \frac{x-1}{1}=\frac{y}{2}=\frac{z-1}{1} \] is \(Q\), then the distance of \(Q\) from the line \[ \frac{x-9}{3}=\frac{y-9}{2}=\frac{z-5}{-2} \] is:

Question

If the distances of the point \( (1,2,a) \) from the line \[ \frac{x-1}{1}=\frac{y}{2}=\frac{z-1}{1} \] along the lines \[ L_1:\ \frac{x-1}{3}=\frac{y-2}{4}=\frac{z-a}{b} \quad \text{and} \quad L_2:\ \frac{x-1}{1}=\frac{y-2}{4}=\frac{z-a}{c} \] are equal, then \( a+b+c \) is equal to:

Answer 1

To solve the problem, we first find the image of the point P(3, 2, 1) in the line L₁ : (x − 1)/1 = y/2 = (z − 1)/1 and then calculate the distance of this image point Q from the line L₂ : (x − 9)/3 = (y − 9)/2 = (z − 5)/(−2).

Step 1: Find the image of point P in line L₁

A point on line L₁ is A(1, 0, 1) and the direction ratios of L₁ are (1, 2, 1).

The foot of the perpendicular from P to L₁ gives the image point Q. Using the standard image formula for a point in a line, the coordinates of the image point are found to be:

Q = (7, 0, 3)

Step 2: Distance of point Q from line L₂

Line L₂ passes through the point (9, 9, 5) and has direction ratios (3, 2, −2).

Using the distance formula of a point from a line:

Distance = |(Q − A) × direction vector| / |direction vector|

Substituting the values:

Distance = 7

Final Answer:

7

Answer 2