Step 1: Understanding the Concept:
The equation of any plane passing through the line of intersection of two planes $P_1 = 0$ and $P_2 = 0$ is given by $P_1 + \lambda P_2 = 0$.
A plane is perpendicular to the XY-plane if its normal vector is perpendicular to the normal vector of the XY-plane.
Step 2: Key Formula or Approach:
1. Equation of family of planes: $(x + y + z - 1) + \lambda(3x + 4y + 5z - 2) = 0$.
2. Find the normal vector $\vec{n}$ of this plane.
3. The XY-plane is $z = 0$, its normal is $\hat{k} = \langle 0, 0, 1 \rangle$.
4. Apply the perpendicularity condition: $\vec{n} \cdot \hat{k} = 0$ to find $\lambda$.
Step 3: Detailed Explanation:
The family of planes is:
\[ (x + y + z - 1) + \lambda(3x + 4y + 5z - 2) = 0 \]
Rearranging terms to group $x, y, z$:
\[ (1 + 3\lambda)x + (1 + 4\lambda)y + (1 + 5\lambda)z - (1 + 2\lambda) = 0 \]
The normal vector to this plane is:
\[ \vec{n} = (1 + 3\lambda)\hat{i} + (1 + 4\lambda)\hat{j} + (1 + 5\lambda)\hat{k} \]
The given plane is perpendicular to the XY-plane. The equation of the XY-plane is $z = 0$, so its normal vector is $\vec{n}_{xy} = \hat{k}$.
For the two planes to be perpendicular, their normal vectors must be orthogonal:
\[ \vec{n} \cdot \vec{n}_{xy} = 0 \]
\[ \langle 1 + 3\lambda, 1 + 4\lambda, 1 + 5\lambda \rangle \cdot \langle 0, 0, 1 \rangle = 0 \]
\[ 0 + 0 + (1 + 5\lambda)(1) = 0 \]
\[ 1 + 5\lambda = 0 \implies \lambda = -\frac{1}{5} \]
Now, substitute $\lambda = -\frac{1}{5}$ back into the equation of the family of planes:
\[ \left(1 + 3\left(-\frac{1}{5}\right)\right)x + \left(1 + 4\left(-\frac{1}{5}\right)\right)y + \left(1 + 5\left(-\frac{1}{5}\right)\right)z - \left(1 + 2\left(-\frac{1}{5}\right)\right) = 0 \]
\[ \left(1 - \frac{3}{5}\right)x + \left(1 - \frac{4}{5}\right)y + (1 - 1)z - \left(1 - \frac{2}{5}\right) = 0 \]
\[ \frac{2}{5}x + \frac{1}{5}y + 0z - \frac{3}{5} = 0 \]
Multiply the entire equation by 5 to clear the denominators:
\[ 2x + y - 3 = 0 \]
Step 4: Final Answer:
The required plane equation is $2x + y - 3 = 0$.