JEE Advanced 2019 Paper 2 Mathematics: Paragraph 2

The normal vector is $(2,3,6)$ with magnitude $\sqrt{4+9+36}=7$. (P) Distance of the origin $=\frac{|0-42|}{7}=6$, matching (4). (Q) For $(0,7,7)$: $\frac{|21+42-42|}{7}=3$, matching (2). (R) Writing the plane as $\frac{x}{21}+\frac{y}{14}+\frac{z}{7}=1$, the intercepts are $21,14,7$, so the volume is $\frac16\cdot21\cdot14\cdot7=343$, matching (6). (S) The $z$-intercept is $7$, matching (5). Note that the distance of the origin ($6$) and the distance of $(0,7,7)$ ($3$) must not be interchanged, and the volume of a tetrahedron with intercepts $p,q,r$ is $\frac16pqr$. The correct matching is P→4, Q→2, R→6, S→5.

With $|\vec n|=7$: (P) $\frac{|8+12+36-42|}{7}=\frac{14}7=2$, matching (1). (Q) $\frac{|14+21+42-42|}{7}=\frac{35}7=5$, matching (3). (R) The distance between parallel planes $2x+3y+6z=42$ and $=84$ is $\frac{|84-42|}{7}=6$, matching (4). (S) The intercepts are $21,14,7$ with sum $42$, and $\frac{42}6=7$, matching (5). The option beginning P→3, Q→1 interchanges the two point distances, while the option ending R→5, S→4 interchanges the plane-to-plane distance and the intercept quantity. The correct matching is P→1, Q→3, R→4, S→5.