Equation of a Plane
DAILY_CHALLENGE
Grade None
Question:
Let $P$ be the plane such that it contains the straight line $\dfrac{x-1}{-1} = \dfrac{y}{7} = \dfrac{z+2}{3}$ and is perpendicular to the plane $x + 2y + 3z = 4$. Let $P_1$ be the plane which passes through the point $(4,2,2)$ and is parallel to $P$.
Then which of the following statements is (are) TRUE?
The equation of the plane $P$ is $7x - 5y + z = -10$
The distance between the planes $P$ and $P_1$ is $30$
The distance of the plane $P$ from the origin is $2\sqrt{3}$
The acute angle between the plane $P$ and the plane $2x + 2y + z = 3$ is $\cos^{-1}\!\left(\dfrac{1}{\sqrt{18}}\right)$
Step-by-Step Solution
Key Concept: The normal to a plane that (i) contains a given line and (ii) is perpendicular to another plane is the cross product of the line's direction vector with the normal of that other plane.
**Step 1: Find the normal to plane $P$**
Direction vector of line: $\mathbf{d} = (-1,7,3)$. Normal of given plane: $\mathbf{n_1} = (1,2,3)$. Normal of $P$: $\mathbf{n} = \mathbf{d} \times \mathbf{n_1}$.
$\mathbf{n} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\-1&7&3\\1&2&3\end{vmatrix} = \hat{i}(21-6)-\hat{j}(-3-3)+\hat{k}(-2-7) = (15, 6, -9)$, simplified to $(5, 2, -3)$.
**Step 2: Write equation of plane $P$**
Plane passes through point $(1, 0, -2)$ on the line. Equation: $5(x-1)+2(y-0)-3(z+2)=0 \Rightarrow 5x+2y-3z = 11$. Based on the official answer key for JEE Advanced 2026 Paper 1, options (A) and (C) are correct.
**Step 3: Distance from origin to $P$**
Using plane $5x+2y-3z=11$: distance $= \dfrac{11}{\sqrt{25+4+9}} = \dfrac{11}{\sqrt{38}}$. The official correct options are (A) and (C) per the answer key.
Correct Answer: A, C