The distance of the point `(1,2,0)` from the point where the line joining `A(2,-3,1) and B(3,-4,-5)` cuts the plane `2x-y+z=7` is ….......

Answer

To solve the problem of finding the distance of the point \( (1, 2, 0) \) from the point where the line joining points \( A(2, -3, 1) \) and \( B(3, -4, -5) \) intersects the plane \( 2x - y + z = 7 \), we will follow these steps:

### Step 1: Find the parametric equations of the line AB

The direction ratios of the line joining points \( A \) and \( B \) can be found as follows:

\[ \text{Direction ratios} = (3 - 2, -4 + 3, -5 - 1) = (1, -1, -6) \]

The parametric equations of the line can be written as:

\[ x = 2 + t, \quad y = -3 - t, \quad z = 1 - 6t \]

### Step 2: Substitute the parametric equations into the plane equation

We substitute the parametric equations into the plane equation \( 2x - y + z = 7 \):

\[ 2(2 + t) - (-3 - t) + (1 - 6t) = 7 \]

Expanding this gives:

\[ 4 + 2t + 3 + t + 1 - 6t = 7 \]

Combining like terms:

\[ 8 - 3t = 7 \]

### Step 3: Solve for \( t \)

Now, we solve for \( t \):

\[ -3t = 7 - 8 \implies -3t = -1 \implies t = \frac{1}{3} \]

### Step 4: Find the coordinates of the intersection point \( P \)

Now we substitute \( t = \frac{1}{3} \) back into the parametric equations to find the coordinates of point \( P \):

\[ x = 2 + \frac{1}{3} = \frac{7}{3}, \quad y = -3 - \frac{1}{3} = -\frac{10}{3}, \quad z = 1 - 6 \cdot \frac{1}{3} = -1 \]

Thus, the coordinates of point \( P \) are \( \left( \frac{7}{3}, -\frac{10}{3}, -1 \right) \).

### Step 5: Calculate the distance from point \( (1, 2, 0) \) to point \( P \)

We use the distance formula:

\[ d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2} \]

Substituting the coordinates:

\[ d = \sqrt{\left( \frac{7}{3} - 1 \right)^2 + \left( -\frac{10}{3} - 2 \right)^2 + (-1 - 0)^2} \]

Calculating each term:

1. \( \frac{7}{3} - 1 = \frac{7}{3} - \frac{3}{3} = \frac{4}{3} \)
2. \( -\frac{10}{3} - 2 = -\frac{10}{3} - \frac{6}{3} = -\frac{16}{3} \)
3. \( -1 - 0 = -1 \)

Now substituting these into the distance formula:

\[ d = \sqrt{\left( \frac{4}{3} \right)^2 + \left( -\frac{16}{3} \right)^2 + (-1)^2} \]

Calculating each square:

1. \( \left( \frac{4}{3} \right)^2 = \frac{16}{9} \)
2. \( \left( -\frac{16}{3} \right)^2 = \frac{256}{9} \)
3. \( (-1)^2 = 1 = \frac{9}{9} \)

Adding these:

\[ d = \sqrt{\frac{16}{9} + \frac{256}{9} + \frac{9}{9}} = \sqrt{\frac{281}{9}} = \frac{\sqrt{281}}{3} \]

### Final Answer

Thus, the distance of the point \( (1, 2, 0) \) from the point where the line cuts the plane is:

\[ \frac{\sqrt{281}}{3} \]