Let "PQR" be a triangle with "R(-1,4,2)" .Suppose "M(2,1,2)" is the mid point of PQ .The distance of the centroid of `/_PQR` from the point of intersection of the lines `(x-2)/(0)=(y)/(2)=(z+3)/(-1)` and `(x-1)/(1)=(y+3)/(-3)=(z+1)/(1)` is

Options

A. 9
B. `sqrt69`
C. `sqrt99`
D. 69

Answer

To solve the problem, we need to find the distance of the centroid of triangle PQR from the point of intersection of two given lines. Let's break this down step by step.

### Step 1: Identify the coordinates of points R and M
We are given:
- Point R = (-1, 4, 2)
- Midpoint M = (2, 1, 2)

### Step 2: Use the midpoint formula to express coordinates of P and Q
Since M is the midpoint of PQ, we can express the coordinates of P and Q as follows:
- Let P = (x1, y1, z1)
- Let Q = (x2, y2, z2)

Using the midpoint formula:
\[ M = \left(\frac{x1 + x2}{2}, \frac{y1 + y2}{2}, \frac{z1 + z2}{2}\right) \]
From the coordinates of M, we have:
\[ \frac{x1 + x2}{2} = 2 \implies x1 + x2 = 4 \]
\[ \frac{y1 + y2}{2} = 1 \implies y1 + y2 = 2 \]
\[ \frac{z1 + z2}{2} = 2 \implies z1 + z2 = 4 \]

### Step 3: Calculate the coordinates of the centroid C of triangle PQR
The centroid C of triangle PQR is given by the formula:
\[ C = \left(\frac{x1 + x2 + x3}{3}, \frac{y1 + y2 + y3}{3}, \frac{z1 + z2 + z3}{3}\right) \]
Substituting the known values:
- \(x3 = -1\)
- \(y3 = 4\)
- \(z3 = 2\)

Thus,
\[ C_x = \frac{4 - 1}{3} = \frac{3}{3} = 1 \]
\[ C_y = \frac{2 + 4}{3} = \frac{6}{3} = 2 \]
\[ C_z = \frac{4 + 2}{3} = \frac{6}{3} = 2 \]
So, the coordinates of the centroid C are (1, 2, 2).

### Step 4: Find the intersection of the two lines
The first line is given by:
\[ \frac{x - 2}{0} = \frac{y}{2} = \frac{z + 3}{-1} = k \]
From this, we can deduce:
- \(x = 2\) (since the denominator is 0)
- \(y = 2k\)
- \(z = -3 - k\)

The second line is given by:
\[ \frac{x - 1}{1} = \frac{y + 3}{-3} = \frac{z + 1}{1} = m \]
From this, we can deduce:
- \(x = 1 + m\)
- \(y = -3m - 3\)
- \(z = m - 1\)

Setting \(x\) from both equations equal:
\[ 2 = 1 + m \implies m = 1 \]

Now substituting \(m = 1\) into the second line's equations:
\[ y = -3(1) - 3 = -6 \]
\[ z = 1 - 1 = 0 \]
Thus, the point of intersection of the two lines is (2, -6, 0).

### Step 5: Calculate the distance between the centroid C and the intersection point
Using the distance formula:
\[ d = \sqrt{(C_x - P_x)^2 + (C_y - P_y)^2 + (C_z - P_z)^2} \]
Substituting the coordinates:
\[ d = \sqrt{(1 - 2)^2 + (2 - (-6))^2 + (2 - 0)^2} \]
\[ = \sqrt{(-1)^2 + (8)^2 + (2)^2} \]
\[ = \sqrt{1 + 64 + 4} = \sqrt{69} \]

### Final Answer
The distance of the centroid of triangle PQR from the point of intersection of the lines is \(\sqrt{69}\).

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