What is the equation of the plane passing through the point (1,1,1) and perpendicular to the line whose direction ratios are (3,2,1)?

Calculation:

Given,

The plane passes through the point \(P(1, 1, 1)\) and is perpendicular to the line with direction ratios \( (3, 2, 1) \).

The general equation of the plane is:

\(Ax + By + Cz = D\), where \( (A, B, C) \) are the direction ratios of the normal to the plane.

Since the plane is perpendicular to the line, the normal vector to the plane is \( (3, 2, 1) \).

Thus, the equation of the plane becomes:

\( 3x + 2y + z = D \).

To find \( D \), substitute the point \( (1, 1, 1) \) into the equation of the plane:

\( 3(1) + 2(1) + 1 = D \), which simplifies to:

\( 3 + 2 + 1 = D \),

\( D = 6 \).

∴ The equation of the plane is \( 3x + 2y + z = 6 \).

Hence, the correct answer is Option 2.