The problem asks us to find the direction ratios of the line of intersection, denoted as S, formed by two given planes.

Plane Equations

The equations of the two planes are:

• Plane 1: \( x + y + z = 1 \)
• Plane 2: \( 2x + 3y - 4z = 8 \)

Finding Direction Ratios of Line of Intersection

The line of intersection S is perpendicular to the normal vectors of both planes. Therefore, the direction ratios of the line S can be found by taking the cross product of the normal vectors of the two planes.

Normal Vectors

The normal vector (\(\vec{n}\)) to a plane defined by the equation \( Ax + By + Cz = D \) is given by the coefficients of x, y, and z, i.e., \(\vec{n} = \begin{pmatrix} A \\ B \\ C \end{pmatrix}\).

• For Plane 1 (\(x + y + z = 1\)), the normal vector is \(\vec{n_1} = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}\).
• For Plane 2 (\(2x + 3y - 4z = 8\)), the normal vector is \(\vec{n_2} = \begin{pmatrix} 2 \\ 3 \\ -4 \end{pmatrix}\).

Calculating the Cross Product

The direction ratios of the line of intersection S are given by \(\vec{d} = \vec{n_1} \times \vec{n_2}\).

Let's calculate the cross product:

Conclusion

The calculated direction ratios of the line S are \((-7, 6, 1)\). Comparing this result with the given options, we find that option 2 matches our calculation.

• Option 1: \((-7, -6, 1)\)
• Option 2: \((-7, 6, 1)\)
• Option 3: \((-6, 5, 1)\)
• Option 4: \((6, 5, 1)\)

Therefore, the correct direction ratios for the line S are \((-7, 6, 1)\).