Let the point, on the line passing through the points `P(1, -2, 3)` and `Q(5, -4, 7)`, farther from the origin and at a distance of 9 units from the point P, be `(alpha, beta, gamma)`. Then `alpha^2 + beta^2 + gamma^2` is equal to

Options

A. 160
B. 155
C. 150
D. 165

Answer

To solve the problem, we need to find the coordinates of the point \( ( \alpha, \beta, \gamma ) \) that lies on the line passing through points \( P(1, -2, 3) \) and \( Q(5, -4, 7) \), is farther from the origin, and is at a distance of 9 units from point \( P \).

### Step 1: Find the direction ratios of the line PQ
The direction ratios of the line passing through points \( P \) and \( Q \) can be calculated as follows:

\[ \text{Direction ratios} = (x_2 - x_1, y_2 - y_1, z_2 - z_1) = (5 - 1, -4 - (-2), 7 - 3) = (4, -2, 4) \]

### Step 2: Parametric equations of the line
Using the direction ratios, we can write the parametric equations of the line:

\[ x = 1 + 4t, \quad y = -2 - 2t, \quad z = 3 + 4t \]

### Step 3: Find the point at a distance of 9 units from point P
We need to find the value of \( t \) such that the distance from point \( P(1, -2, 3) \) to the point \( R(x, y, z) \) is 9 units. The distance formula gives us:

\[ \sqrt{(x - 1)^2 + (y + 2)^2 + (z - 3)^2} = 9 \]

Squaring both sides, we have:

\[ (x - 1)^2 + (y + 2)^2 + (z - 3)^2 = 81 \]

Substituting the parametric equations into the distance equation:

\[ (4t)^2 + (-2 - 2t + 2)^2 + (4t)^2 = 81 \]

This simplifies to:

\[ (4t)^2 + (-2t)^2 + (4t)^2 = 81 \]

\[ 16t^2 + 4t^2 + 16t^2 = 81 \]

\[ 36t^2 = 81 \]

### Step 4: Solve for \( t \)
Dividing both sides by 36:

\[ t^2 = \frac{81}{36} = \frac{9}{4} \]

Taking the square root:

\[ t = \frac{3}{2} \quad \text{or} \quad t = -\frac{3}{2} \]

### Step 5: Find the coordinates of points R
1. For \( t = \frac{3}{2} \):

\[ x = 1 + 4 \cdot \frac{3}{2} = 1 + 6 = 7 \]
\[ y = -2 - 2 \cdot \frac{3}{2} = -2 - 3 = -5 \]
\[ z = 3 + 4 \cdot \frac{3}{2} = 3 + 6 = 9 \]

So, one point is \( R(7, -5, 9) \).

2. For \( t = -\frac{3}{2} \):

\[ x = 1 + 4 \cdot -\frac{3}{2} = 1 - 6 = -5 \]
\[ y = -2 - 2 \cdot -\frac{3}{2} = -2 + 3 = 1 \]
\[ z = 3 + 4 \cdot -\frac{3}{2} = 3 - 6 = -3 \]

So, the other point is \( R(-5, 1, -3) \).

### Step 6: Determine which point is farther from the origin
Now we calculate the distances from the origin \( (0, 0, 0) \):

1. For \( R(7, -5, 9) \):

\[ \sqrt{7^2 + (-5)^2 + 9^2} = \sqrt{49 + 25 + 81} = \sqrt{155} \]

2. For \( R(-5, 1, -3) \):

\[ \sqrt{(-5)^2 + 1^2 + (-3)^2} = \sqrt{25 + 1 + 9} = \sqrt{35} \]

Since \( \sqrt{155} > \sqrt{35} \), the point \( (7, -5, 9) \) is farther from the origin.

### Step 7: Calculate \( \alpha^2 + \beta^2 + \gamma^2 \)
Now, we calculate:

\[ \alpha^2 + \beta^2 + \gamma^2 = 7^2 + (-5)^2 + 9^2 = 49 + 25 + 81 = 155 \]

Thus, the final answer is:

\[ \alpha^2 + \beta^2 + \gamma^2 = 155 \]