Vector Equation Of A Plane

The relevant standard is the normal form of a plane: \vec{r} \cdot \hat{n} = p, where \hat{n} is the unit normal and p is the perpendicular distance from the origin. This form is a fundamental JEE Advanced representation directly linking distance and orientation. Here \hat{n} = (1/3)(2i + 2j + k), which is indeed a unit vector since \sqrt{2^2 + 2^2 + 1^2}/3 = 3/3 = 1, and p = 3. Thus \vec{r} \cdot (1/3)(2i + 2j + k) = 3, and multiplying both sides by 3 gives the tidy Cartesian-style form \vec{r} \cdot (2i + 2j + k) = 9. Option \vec{r} \cdot (2i + 2j + k) = 3 forgets to scale the right side when clearing the 1/3. Option = 1 mishandles both the distance and the normal. The last option uses a non-unit, wrong normal direction. This applies the normal-form plane theorem, which is special because the coefficient on the right side carries direct geometric meaning only when the normal is a genuine unit vector; verifying that \hat{n} has length one is therefore an essential first step before reading off the distance. Many errors arise from treating an unnormalized normal as if it were already a unit vector. Plausibility check: dividing the result back gives \vec{r} \cdot \hat{n} = 9/3 = 3 = p, recovering the stated perpendicular distance correctly, and since p is positive the plane sits a real, nonzero distance from the origin rather than passing through it, which matches the problem's intent.