Let the equation of a line is `(x-2)/(1)=(y-3)/(2)=(z-4)/(3)`. An insect starts flying from `P(1, 3, 2)` in a straight line meeting the given line at a point R(a, b, c) and then goes to the point Q (6, 7, 5) in a straight line such that PR is perpendicular to RQ. Then the least value of `7(a+b+c)` is equal to

Options

A. 105
B. 45
C. 10
D. 7

Answer

To solve the problem step by step, we will follow the given instructions and derive the least value of \( 7(a + b + c) \).

### Step 1: Parametrize the line
The equation of the line is given as:
\[ \frac{x - 2}{1} = \frac{y - 3}{2} = \frac{z - 4}{3} \]
Let \( t \) be the parameter. Then we can express the coordinates of any point \( R(a, b, c) \) on the line as:
\[ x = 1t + 2 = t + 2 \]
\[ y = 2t + 3 \]
\[ z = 3t + 4 \]
Thus, the coordinates of point \( R \) can be written as:
\[ R(t + 2, 2t + 3, 3t + 4) \]

### Step 2: Find direction ratios of \( PR \) and \( RQ \)
The coordinates of point \( P \) are \( (1, 3, 2) \) and point \( Q \) are \( (6, 7, 5) \).

The direction ratios of vector \( PR \) are:
\[ PR = R - P = (t + 2 - 1, 2t + 3 - 3, 3t + 4 - 2) = (t + 1, 2t, 3t + 2) \]

The direction ratios of vector \( RQ \) are:
\[ RQ = Q - R = (6 - (t + 2), 7 - (2t + 3), 5 - (3t + 4)) = (4 - t, 4 - 2t, 1 - 3t) \]

### Step 3: Set up the perpendicular condition
Since \( PR \) is perpendicular to \( RQ \), their dot product must equal zero:
\[ (t + 1)(4 - t) + (2t)(4 - 2t) + (3t + 2)(1 - 3t) = 0 \]

### Step 4: Expand and simplify the equation
Expanding the dot product:
1. \( (t + 1)(4 - t) = 4t + 4 - t^2 - t = -t^2 + 3t + 4 \)
2. \( (2t)(4 - 2t) = 8t - 4t^2 \)
3. \( (3t + 2)(1 - 3t) = 3t - 9t^2 + 2 - 6t = -9t^2 - 3t + 2 \)

Combining these:
\[ -t^2 + 3t + 4 + 8t - 4t^2 - 9t^2 - 3t + 2 = 0 \]
This simplifies to:
\[ -14t^2 + 8t + 6 = 0 \]
Dividing by -2:
\[ 7t^2 - 4t - 3 = 0 \]

### Step 5: Solve the quadratic equation
Using the quadratic formula:
\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{4 \pm \sqrt{(-4)^2 - 4 \cdot 7 \cdot (-3)}}{2 \cdot 7} \]
\[ = \frac{4 \pm \sqrt{16 + 84}}{14} = \frac{4 \pm \sqrt{100}}{14} = \frac{4 \pm 10}{14} \]
This gives:
\[ t = \frac{14}{14} = 1 \quad \text{or} \quad t = \frac{-6}{14} = -\frac{3}{7} \]

### Step 6: Find \( a + b + c \)
Substituting \( t = 1 \):
\[ a = 1 + 2 = 3, \quad b = 2(1) + 3 = 5, \quad c = 3(1) + 4 = 7 \]
Thus, \( a + b + c = 3 + 5 + 7 = 15 \).

Substituting \( t = -\frac{3}{7} \):
\[ a = -\frac{3}{7} + 2 = \frac{11}{7}, \quad b = 2(-\frac{3}{7}) + 3 = \frac{15}{7}, \quad c = 3(-\frac{3}{7}) + 4 = \frac{9}{7} \]
Thus, \( a + b + c = \frac{11}{7} + \frac{15}{7} + \frac{9}{7} = \frac{35}{7} = 5 \).

### Step 7: Calculate the least value of \( 7(a + b + c) \)
For \( t = 1 \):
\[ 7(a + b + c) = 7 \times 15 = 105 \]
For \( t = -\frac{3}{7} \):
\[ 7(a + b + c) = 7 \times 5 = 35 \]

### Conclusion
The least value of \( 7(a + b + c) \) is:
\[ \boxed{35} \]