Let d = the distance of point `A(4,1,1)` from the line of intersection of the planes `x+y+z=4 and x-2y-z=4`, then `14d^(2)//3` = __________

Answer

To find the distance \( d \) of the point \( A(4, 1, 1) \) from the line of intersection of the planes \( x + y + z = 4 \) and \( x - 2y - z = 4 \), we can follow these steps:

### Step 1: Find the direction vector of the line of intersection
To find the direction vector of the line of intersection of the two planes, we can take the cross product of the normal vectors of the planes.

The normal vector of the first plane \( x + y + z = 4 \) is \( \mathbf{n_1} = (1, 1, 1) \) and the normal vector of the second plane \( x - 2y - z = 4 \) is \( \mathbf{n_2} = (1, -2, -1) \).

Now, we compute the cross product \( \mathbf{n_1} \times \mathbf{n_2} \):
\[ \mathbf{d} = \mathbf{n_1} \times \mathbf{n_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 1 & 1 \\ 1 & -2 & -1 \end{vmatrix} \]
Calculating the determinant:
\[ \mathbf{d} = \mathbf{i} \begin{vmatrix} 1 & 1 \\ -2 & -1 \end{vmatrix} - \mathbf{j} \begin{vmatrix} 1 & 1 \\ 1 & -1 \end{vmatrix} + \mathbf{k} \begin{vmatrix} 1 & 1 \\ 1 & -2 \end{vmatrix} \]
\[ = \mathbf{i} (1 \cdot -1 - 1 \cdot -2) - \mathbf{j} (1 \cdot -1 - 1 \cdot 1) + \mathbf{k} (1 \cdot -2 - 1 \cdot 1) \]
\[ = \mathbf{i} (-1 + 2) - \mathbf{j} (-1 - 1) + \mathbf{k} (-2 - 1) \]
\[ = \mathbf{i} (1) + \mathbf{j} (2) - \mathbf{k} (3) = (1, 2, -3) \]

### Step 2: Find a point on the line of intersection
To find a point on the line of intersection, we can set \( z = 0 \) and solve the system of equations:
1. \( x + y + 0 = 4 \) → \( x + y = 4 \)
2. \( x - 2y - 0 = 4 \) → \( x - 2y = 4 \)

From the first equation, \( y = 4 - x \). Substituting into the second equation:
\[ x - 2(4 - x) = 4 \]
\[ x - 8 + 2x = 4 \]
\[ 3x = 12 \implies x = 4 \]
Substituting \( x = 4 \) back into \( y = 4 - x \):
\[ y = 4 - 4 = 0 \]
Thus, a point on the line is \( (4, 0, 0) \).

### Step 3: Find the distance from point A to the line
Now, we need to find the distance from point \( A(4, 1, 1) \) to the line defined by the point \( (4, 0, 0) \) and direction vector \( (1, 2, -3) \).

The distance \( d \) from a point \( P(x_1, y_1, z_1) \) to a line defined by point \( Q(x_0, y_0, z_0) \) and direction vector \( \mathbf{v}(a, b, c) \) is given by:
\[ d = \frac{| \mathbf{PQ} \cdot (\mathbf{v} \times \mathbf{PQ}) |}{|\mathbf{v}|} \]
where \( \mathbf{PQ} = (x_1 - x_0, y_1 - y_0, z_1 - z_0) \).

Calculating \( \mathbf{PQ} \):
\[ \mathbf{PQ} = (4 - 4, 1 - 0, 1 - 0) = (0, 1, 1) \]

Now, we compute \( \mathbf{v} \times \mathbf{PQ} \):
\[ \mathbf{v} = (1, 2, -3) \]
\[ \mathbf{v} \times \mathbf{PQ} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & -3 \\ 0 & 1 & 1 \end{vmatrix} \]
Calculating the determinant:
\[ = \mathbf{i} \begin{vmatrix} 2 & -3 \\ 1 & 1 \end{vmatrix} - \mathbf{j} \begin{vmatrix} 1 & -3 \\ 0 & 1 \end{vmatrix} + \mathbf{k} \begin{vmatrix} 1 & 2 \\ 0 & 1 \end{vmatrix} \]
\[ = \mathbf{i} (2 \cdot 1 - (-3) \cdot 1) - \mathbf{j} (1 \cdot 1 - 0 \cdot -3) + \mathbf{k} (1 \cdot 1 - 0 \cdot 2) \]
\[ = \mathbf{i} (2 + 3) - \mathbf{j} (1) + \mathbf{k} (1) \]
\[ = 5\mathbf{i} - 1\mathbf{j} + 1\mathbf{k} = (5, -1, 1) \]

Now, we calculate the magnitude of \( \mathbf{v} \):
\[ |\mathbf{v}| = \sqrt{1^2 + 2^2 + (-3)^2} = \sqrt{1 + 4 + 9} = \sqrt{14} \]

Now, we compute \( | \mathbf{PQ} \cdot (\mathbf{v} \times \mathbf{PQ}) | \):
\[ \mathbf{PQ} \cdot (\mathbf{v} \times \mathbf{PQ}) = (0, 1, 1) \cdot (5, -1, 1) = 0 \cdot 5 + 1 \cdot (-1) + 1 \cdot 1 = -1 + 1 = 0 \]

Thus, the distance \( d \) is:
\[ d = \frac{|0|}{\sqrt{14}} = 0 \]

### Step 4: Calculate \( \frac{14d^2}{3} \)
Since \( d = 0 \):
\[ \frac{14d^2}{3} = \frac{14 \cdot 0^2}{3} = 0 \]

### Final Answer
\[ \frac{14d^2}{3} = 0 \]