Consider three planes P₁ : x - y + z = 1, P₂ : x + y - z = 1, P₃ : x - 3y + 3z = 2.
Let L₁, L₂, L₃ b

Q: Why do we use the cross product to find the line of intersection? Can't we just solve the equations?

You absolutely can solve the equations, but the cross product is a powerful geometric shortcut! Since the line of intersection lies on both planes, it must be perpendicular to the normal vector of the first plane AND the normal vector of the second plane. The cross product $\vec{n}_1 \times \vec{n}_2$ gives us exactly that direction vector. It is the 'DNA' of the line's orientation.

Q: Wait, why do we reject the negative root or get a contradiction like $1=2$?

That contradiction is not a mistake—it is the answer! When you solve a system of linear equations and arrive at a statement like $1=2$, it means the system is inconsistent. Geometrically, this tells us that there is no single point $(x, y, z)$ where all three planes meet. They form a triangular prism, which is a beautiful, valid geometric configuration.

Q: How do I know if the lines are parallel just by looking at the direction vectors?

Two vectors $\vec{u}$ and $\vec{v}$ are parallel if one is a scalar multiple of the other, i.e., $\vec{u} = k\vec{v}$. If you calculate your direction vectors and find they are all proportional to $(0, 1, 1)$, then they are all pointing in the same direction, making the lines parallel.

Question

Consider three planes $P_1 : x - y + z = 1$, $P_2 : x + y - z = 1$, $P_3 : x - 3y + 3z = 2$.
Let $L_1, L_2, L_3$ be the lines of intersection of the planes $P_2$ and $P_3$, $P_3$ and $P_1$, $P_1$ and $P_2$, respectively.
STATEMENT-1 : At least two of the lines $L_1, L_2$ and $L_3$ are non-parallel and
STATEMENT-2 : The three planes do not have a common point.

Options:

(A) STATEMENT - 1 is True, STATEMENT - 2 is True; STATEMENT - 2 is a correct explanation for STATEMENT - 1

(B) STATEMENT - 1 is True, STATEMENT - 2 is True; STATEMENT - 2 is NOT a correct explanation for STATEMENT - 1

(C) STATEMENT - 1 is True, STATEMENT - 2 is False

(D) STATEMENT - 1 is False, STATEMENT - 2 is True

Sigma Percentile · Accepted Answer

### The Geometry of the Prism: A Journey into 3D Space

Welcome, future engineers. Today, we are not just solving a system of linear equations; we are architects of 3D space. We are looking at three planes, $P_1$, $P_2$, and $P_3$, and we want to understand how they interact.

Do they meet at a single point, or do they form a prism? Let us peel back the layers of this problem.

### Phase 1: The DNA of a Plane

Every plane is defined by its normal vector—the vector that stands perfectly perpendicular to its surface. Think of it as the plane's orientation. For our given planes:

$$P_1 : x - y + z = 1$$
$$P_2 : x + y - z = 1$$
$$P_3 : x - 3y + 3z = 2$$

We extract the normal vectors directly from the coefficients of $x$, $y$, and $z$:

$$\vec{n}_1 = (1, -1, 1)$$
$$\vec{n}_2 = (1, 1, -1)$$
$$\vec{n}_3 = (1, -3, 3)$$

These vectors are the keys to the kingdom. If we want to know the direction of the line where two planes meet, we need a vector that is perpendicular to both of their normals. This is the fundamental definition of the cross product.

### Phase 2: The Intersection Lines

Let us find the direction of the line $L_1$, which is the intersection of $P_2$ and $P_3$. The direction vector $\vec{d}_1$ must be $\vec{n}_2 	imes \vec{n}_3$. We compute this using the determinant:

$$\vec{d}_1 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & 1 & -1 \ 1 & -3 & 3 \end{vmatrix}$$

Expanding this, we get $\hat{i}(3 - 3) - \hat{j}(3 - (-1)) + \hat{k}(-3 - 1) = (0, -4, -4)$. We can simplify this direction vector by dividing by $-4$, giving us the direction $(0, 1, 1)$.

Now, let us look at $L_2$ (the intersection of $P_3$ and $P_1$) and $L_3$ (the intersection of $P_1$ and $P_2$). Following the same logic:

$$\vec{d}_2 = \vec{n}_3 	imes \vec{n}_1 = (0, 2, 2) \parallel (0, 1, 1)$$
$$\vec{d}_3 = \vec{n}_1 	imes \vec{n}_2 = (0, 2, 2) \parallel (0, 1, 1)$$

Look at that! All three lines of intersection are parallel to the vector $(0, 1, 1)$. This immediately tells us that Statement 1, which claims at least two lines are non-parallel, is fundamentally false. They are all marching in the same direction.

### Phase 3: The Prism Mystery

Now, we address Statement 2: Do the three planes have a common point? To find out, we attempt to solve the system of equations simultaneously:

1) $x - y + z = 1$
2) $x + y - z = 1$
3) $x - 3y + 3z = 2$

Let us add equations (1) and (2). The $y$ and $z$ terms cancel out beautifully:

$$(x - y + z) + (x + y - z) = 1 + 1$$
$$2x = 2 \implies x = 1$$

Substituting $x = 1$ into equation (1), we get $1 - y + z = 1$, which simplifies to $y = z$. Now, we take these findings—$x = 1$ and $y = z$—and substitute them into equation (3):

$$1 - 3(z) + 3(z) = 2$$
$$1 = 2$$

This is the moment of truth. We have arrived at a mathematical contradiction. The system has no solution.

Geometrically, this means the three planes never intersect at a single common point. They form a triangular prism, where the lines of intersection are parallel edges of the prism. Thus, **Statement 2 is true**.

### Conclusion

We have navigated the geometry of planes and lines. We saw how the cross product reveals the hidden orientation of intersection lines, and how algebraic contradictions reveal the physical reality of a prism.

Remember, in JEE Advanced, a contradiction is not a failure—it is a geometric conclusion. Keep practicing, keep visualizing, and keep falling in love with the logic of the universe.

Select Answer:

Visualized Solution

The 3D Geometry of Three Planes

Normal Vectors of the Planes

Direction of Intersection Line $L_{1}$

Calculating $d_{1}$

Direction of Line $L_{2}$

Direction of Line $L_{3}$

Evaluating Statement 1

Checking for a Common Point

Solving the System (Part 1)

Solving the System (Part 2)

Final Conclusion

The Sigma Insight: Equation of a Plane

The Geometry of the Prism

Phase 1

Phase 2

Phase 3

Conclusion

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