To solve the problem, we need to find the coordinates of a point on the plane \(2x + y - 5z = 0\) that is \(2\sqrt{11}\) units away from the line of intersection of the planes \(2x + y - 5z = 0\) and \(4x - 3y + 7z = 0\).
### Step-by-Step Solution:
1. **Identify the planes**:
The two planes given are:
- Plane 1: \(P_1: 2x + y - 5z = 0\)
- Plane 2: \(P_2: 4x - 3y + 7z = 0\)
2. **Find the direction ratios of the line of intersection**:
The direction ratios of the line of intersection of two planes can be found using the cross product of the normal vectors of the planes.
- Normal vector of Plane 1: \(\vec{n_1} = (2, 1, -5)\)
- Normal vector of Plane 2: \(\vec{n_2} = (4, -3, 7)\)
The direction ratios of the line of intersection \(\vec{d}\) can be calculated as:
\[
\vec{d} = \vec{n_1} \times \vec{n_2}
\]
\[
\vec{d} = \begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
2 & 1 & -5 \\
4 & -3 & 7
\end{vmatrix}
\]
Calculating the determinant:
\[
\vec{d} = \hat{i} \left(1 \cdot 7 - (-5)(-3)\right) - \hat{j} \left(2 \cdot 7 - (-5)(4)\right) + \hat{k} \left(2 \cdot (-3) - 1 \cdot 4\right)
\]
\[
= \hat{i} (7 - 15) - \hat{j} (14 + 20) + \hat{k} (-6 - 4)
\]
\[
= \hat{i} (-8) - \hat{j} (34) + \hat{k} (-10)
\]
Thus, the direction ratios are \((-8, -34, -10)\).
3. **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 equations of the planes:
- From Plane 1: \(2x + y = 0 \Rightarrow y = -2x\)
- From Plane 2: \(4x - 3y = 0 \Rightarrow 4x - 3(-2x) = 0 \Rightarrow 4x + 6x = 0 \Rightarrow 10x = 0 \Rightarrow x = 0\)
Substituting \(x = 0\) into \(y = -2x\):
\[
y = 0
\]
Thus, a point on the line of intersection is \((0, 0, 0)\).
4. **Find a point on Plane 1 at a distance of \(2\sqrt{11}\)**:
The distance \(d\) from a point \((x_1, y_1, z_1)\) to a line given by a point \((x_0, y_0, z_0)\) and direction ratios \((a, b, c)\) is given by:
\[
d = \frac{|(x_1 - x_0)(b) - (y_1 - y_0)(a) + (z_1 - z_0)(c)|}{\sqrt{a^2 + b^2 + c^2}}
\]
Let \((x, y, z)\) be a point on Plane 1. We need to satisfy:
\[
2\sqrt{11} = \frac{|(x - 0)(-34) - (y - 0)(-8) + (z - 0)(-10)|}{\sqrt{(-8)^2 + (-34)^2 + (-10)^2}}
\]
Calculate the denominator:
\[
\sqrt{64 + 1156 + 100} = \sqrt{1320} = 2\sqrt{330}
\]
Thus, we have:
\[
2\sqrt{11} = \frac{|(-34)x + 8y - 10z|}{2\sqrt{330}}
\]
Simplifying gives:
\[
|(-34)x + 8y - 10z| = 2\sqrt{11} \cdot 2\sqrt{330} = 4\sqrt{3630}
\]
5. **Find coordinates satisfying Plane 1**:
We can try different values for \(x\) and \(y\) to find \(z\) such that the point lies on Plane 1 and satisfies the distance condition.
After testing various combinations, we find that the point \((6, -2, 2)\) satisfies both the plane equation and the distance requirement.
### Final Answer:
The coordinates of the point on the plane \(2x + y - 5z = 0\) that is \(2\sqrt{11}\) units away from the line of intersection are \((6, -2, 2)\).
