The distance of the point `(3,0,5)` from the line `x-2y+2z-4=0=x+3z-11` is

Explanation

To find the distance of the point \( P(3, 0, 5) \) from the line defined by the equations \( x - 2y + 2z - 4 = 0 \) and \( x + 3z - 11 = 0 \), we can follow these steps:

### Step 1: Find the direction ratios of the line
The equations of the line can be rewritten in parametric form. We have:
1. \( x - 2y + 2z = 4 \)
2. \( x + 3z = 11 \)

From the second equation, we can express \( x \) in terms of \( z \):
\[ x = 11 - 3z \]

Substituting this into the first equation:
\[ (11 - 3z) - 2y + 2z = 4 \]
\[ 11 - 3z - 2y + 2z = 4 \]
\[ -2y - z = 4 - 11 \]
\[ -2y - z = -7 \quad \Rightarrow \quad 2y + z = 7 \quad \Rightarrow \quad z = 7 - 2y \]

Now substituting \( z \) back into the expression for \( x \):
\[ x = 11 - 3(7 - 2y) = 11 - 21 + 6y = 6y - 10 \]

Thus, we can express the line in parametric form:
\[ x = 6t - 10, \quad y = t, \quad z = 7 - 2t \]
where \( t \) is a parameter. The direction ratios of the line are \( (6, 1, -2) \).

### Step 2: Identify a point on the line
To find a specific point on the line, we can set \( t = 0 \):
\[ x = 6(0) - 10 = -10, \quad y = 0, \quad z = 7 - 2(0) = 7 \]
Thus, a point \( A \) on the line is \( A(-10, 0, 7) \).

### Step 3: Find the vector from point A to point P
The vector \( \vec{AC} \) from point \( A(-10, 0, 7) \) to point \( P(3, 0, 5) \) is:
\[ \vec{AC} = P - A = (3 - (-10), 0 - 0, 5 - 7) = (3 + 10, 0, -2) = (13, 0, -2) \]

### Step 4: Find the cross product of \( \vec{AC} \) and the direction vector \( \vec{b} \)
The direction vector \( \vec{b} \) is \( (6, 1, -2) \). We calculate the cross product \( \vec{AC} \times \vec{b} \):
\[ \vec{AC} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 13 & 0 & -2 \\ 6 & 1 & -2 \end{vmatrix} \]
Calculating the determinant:
\[ = \hat{i} \begin{vmatrix} 0 & -2 \\ 1 & -2 \end{vmatrix} - \hat{j} \begin{vmatrix} 13 & -2 \\ 6 & -2 \end{vmatrix} + \hat{k} \begin{vmatrix} 13 & 0 \\ 6 & 1 \end{vmatrix} \]
\[ = \hat{i} (0 \cdot -2 - (-2) \cdot 1) - \hat{j} (13 \cdot -2 - 6 \cdot -2) + \hat{k} (13 \cdot 1 - 0 \cdot 6) \]
\[ = \hat{i} (0 + 2) - \hat{j} (-26 + 12) + \hat{k} (13) \]
\[ = 2\hat{i} + 14\hat{j} + 13\hat{k} \]
Thus, \( \vec{AC} \times \vec{b} = (2, 14, 13) \).

### Step 5: Find the magnitude of the cross product
The magnitude of \( \vec{AC} \times \vec{b} \) is:
\[ |\vec{AC} \times \vec{b}| = \sqrt{2^2 + 14^2 + 13^2} = \sqrt{4 + 196 + 169} = \sqrt{369} \]

### Step 6: Find the magnitude of the direction vector \( \vec{b} \)
The magnitude of \( \vec{b} \) is:
\[ |\vec{b}| = \sqrt{6^2 + 1^2 + (-2)^2} = \sqrt{36 + 1 + 4} = \sqrt{41} \]

### Step 7: Calculate the distance from point P to the line
The distance \( d \) from point \( P \) to the line is given by:
\[ d = \frac{|\vec{AC} \times \vec{b}|}{|\vec{b}|} = \frac{\sqrt{369}}{\sqrt{41}} = \sqrt{\frac{369}{41}} = \sqrt{9} = 3 \]

Thus, the distance of the point \( (3, 0, 5) \) from the line is \( \boxed{3} \).