The equation of plane through intersection of the planes \(x+2 y+z-1=0\) and \(2 x+y+3 z-2=0\) and perpendicular to plane \(x+y+z=1\) is (1) \(x+4 y-3 z+1=0\) (2) \(x-4 y+3 z-1=0\) (3) \(x-4 y+3 z=0\) (4) \(x+4 y-3 z=0\)

A plane in three-dimensional geometry can be uniquely identified by its equation. This equation can take several forms, but one of the most common is the general form, given by: \( Ax + By + Cz + D = 0 \) where \(A\), \(B\), and \(C\) are the coefficients representing the normal vector to the plane, and \(D\) is the constant term.
In our problem, we need to find a plane that intersects two given planes, \(x + 2y + z - 1 = 0\) and \(2x + y + 3z - 2 = 0\).
The equation of a plane passing through the intersection of two planes can be represented using a parameter \(\backslash lambda\): \((x + 2y + z - 1) + \backslash lambda (2x + y + 3z - 2) = 0 \).
Expand and simplify this to combine like terms and express it in the general form. This will give you a linear combination of the two original planes.

Two planes are perpendicular if their normal vectors are orthogonal to each other. This means that the dot product of their normal vectors must be zero.
The normal vector of the plane \(x + y + z = 1\) is \( (1, 1, 1) \).
The normal for our plane (from the intersection equation) is \ where the coefficients are: \((1 + 2\backslash lambda, 2 + \backslash lambda, 1 + 3\backslash lambda) \).
For the given plane to be perpendicular to \((1, 1, 1)\), we need: \( (1 + 2\backslash lambda) \cdot 1 + (2 + \backslash lambda) \cdot 1 + (1 + 3\backslash lambda) \cdot 1 = 0 \).
This condition will help us determine the value of \(\backslash lambda\), solving for which we get \(\backslash lambda = -\backslash frac{2}{3}\).

At the intersection of two planes, a line is formed. When we add the condition that the resulting plane is perpendicular to a third plane, we can find a specific plane.
Using the value of \(\backslash lambda = -\backslash frac{2}{3}\), we substitute it back into our plane equation:
\[ (1 + 2(-\backslash frac{2}{3}))x + (2 - \backslash frac{2}{3})y + (1 + 3(-\backslash frac{2}{3}))z - (1 - 2 \cdot \frac{2}{3})= 0 \]
Multiplying through by 3 to clear the fractions, we get:
\[ -x + 4y - 3z + 1 = 0 \]
Therefore, the final equation of the plane is: \( x - 4y + 3z - 1 = 0 \), which confirms the correctness of the intersection condition with the given planes.