The distance of the point `B(i+2j+3k)` from the line which is passing through `A(4i +2j+2k)` and which is parallel to the vector `vecC =2i+3j+6k` is .......

Answer

To find the distance of the point \( B(1i + 2j + 3k) \) from the line passing through the point \( A(4i + 2j + 2k) \) and parallel to the vector \( \vec{C} = 2i + 3j + 6k \), we can follow these steps:

### Step 1: Identify the Points and Vectors
- Point \( A = (4, 2, 2) \)
- Point \( B = (1, 2, 3) \)
- Direction vector of the line \( \vec{C} = (2, 3, 6) \)

### Step 2: Find the Vector \( \vec{AB} \)
The vector \( \vec{AB} \) can be calculated as:
\[ \vec{AB} = B - A = (1 - 4)i + (2 - 2)j + (3 - 2)k = -3i + 0j + 1k = -3i + k \]

### Step 3: Find the Projection of \( \vec{AB} \) onto \( \vec{C} \)
To find the projection of \( \vec{AB} \) onto \( \vec{C} \), we use the formula:
\[ \text{proj}_{\vec{C}} \vec{AB} = \frac{\vec{AB} \cdot \vec{C}}{\|\vec{C}\|^2} \vec{C} \]

First, calculate \( \vec{AB} \cdot \vec{C} \):
\[ \vec{AB} \cdot \vec{C} = (-3)(2) + (0)(3) + (1)(6) = -6 + 0 + 6 = 0 \]

Next, calculate the magnitude of \( \vec{C} \):
\[ \|\vec{C}\| = \sqrt{2^2 + 3^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7 \]

Now, calculate \( \|\vec{C}\|^2 \):
\[ \|\vec{C}\|^2 = 49 \]

Thus, the projection is:
\[ \text{proj}_{\vec{C}} \vec{AB} = \frac{0}{49} \vec{C} = 0 \]

### Step 4: Calculate the Length of \( \vec{AB} \)
The length of \( \vec{AB} \) is given by:
\[ \|\vec{AB}\| = \sqrt{(-3)^2 + 0^2 + 1^2} = \sqrt{9 + 0 + 1} = \sqrt{10} \]

### Step 5: Calculate the Distance from Point \( B \) to the Line
Since the projection of \( \vec{AB} \) onto \( \vec{C} \) is zero, the distance from point \( B \) to the line is simply the length of \( \vec{AB} \):
\[ \text{Distance} = \|\vec{AB}\| = \sqrt{10} \]

### Final Answer
The distance of the point \( B(1i + 2j + 3k) \) from the line is \( \sqrt{10} \).

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