Let S be the set of all real values of λ such that a plane passing through the points (-λ², 1, 1), (

Q: Why did we choose point A as the common origin for our vectors? Could we have picked D?

That is a brilliant question! You absolutely could have picked D. The choice of the 'origin' point for your vectors is arbitrary. As long as you form three vectors that lie within the plane (like $\vec{DA}, \vec{DB}, \vec{DC}$), the scalar triple product will still be zero. The math is invariant to your choice of starting point!

Q: Wait, why can we just divide by $(1+\lambda^2)^2$? What if it equals zero?

I love that you are checking for division-by-zero errors! In JEE problems, always check the domain. Since $\lambda$ is a real number, $\lambda^2$ is always non-negative. Thus, $1+\lambda^2$ is at minimum 1. It can never be zero, so it is perfectly safe to divide. Always keep this 'domain check' habit—it saves lives in competitive exams!

Q: Is there a faster way to solve this without using the scalar triple product?

You could find the equation of the plane using the first three points and then substitute the fourth point into that equation. However, that often leads to messy algebraic expressions involving $\lambda$. The scalar triple product is the 'elegant' path because it handles the coplanarity condition in one clean, symmetric sweep.

Q: Why did we expand along the third column? Does it matter?

It doesn't change the result, but it changes the effort! Expanding along a row or column with the most zeros is a classic 'pro-tip.' It minimizes the number of terms you have to calculate, reducing the chance of a silly arithmetic error. Always scan for zeros before you start expanding!

Q: I got $\lambda^2 = 3$, but I forgot the negative root. Is that a problem?

In the context of JEE, yes! The question asks for the set $S$ of all real values. If you miss the negative root, you lose half the solution. Whenever you see $\lambda^2 = k$, always pause and remind yourself: 'Is there a negative counterpart?' It is a small detail that separates the top rankers from the rest.

Question

Let $S$ be the set of all real values of $\lambda$ such that a plane passing through the points $(-\lambda^2, 1, 1)$, $(1, -\lambda^2, 1)$ and $(1, 1, -\lambda^2)$ also passes through the point $(-1, -1, 1)$. Then $S$ is equal to :

Options:

(A) $\{\sqrt{3}\}$

(B) $\{\sqrt{3}, -\sqrt{3}\}$

(C) $\{1, -1\}$

(D) $\{3, -3\}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing in a vast, three-dimensional coordinate system. You are given four points: $A(-\lambda^2, 1, 1)$, $B(1, -\lambda^2, 1)$, $C(1, 1, -\lambda^2)$, and $D(-1, -1, 1)$.

The problem asks us to find the values of $\lambda$ such that these four points sit perfectly on the same flat surface—a plane. This is not just an algebraic exercise; it is a test of your ability to visualize the constraints of space.

### The Power of Vectors

When we talk about points being coplanar, we are essentially saying that the vectors connecting them are 'trapped' in the same flat world. If we pick one point, say $A$, as our anchor, we can define three vectors: $\vec{AB}$, $\vec{AC}$, and $\vec{AD}$.

If these three vectors lie in the same plane, they cannot span any volume. In the language of linear algebra, the volume of the parallelepiped formed by these vectors must be zero. This is the physical soul of the scalar triple product: $[\vec{AB} \quad \vec{AC} \quad \vec{AD}] = 0$.

Let us calculate these vectors. By subtracting the position vector of $A$ from $B$, $C$, and $D$, we get:

$$\vec{AB} = (1 + \lambda^2)\hat{i} - (1 + \lambda^2)\hat{j} + 0\hat{k}$$

$$\vec{AC} = (1 + \lambda^2)\hat{i} + 0\hat{j} - (1 + \lambda^2)\hat{k}$$

$$\vec{AD} = (\lambda^2 - 1)\hat{i} - 2\hat{j} + 0\hat{k}$$

### The Determinant Dance

Now, we assemble these into a determinant. We are looking for:

$$\begin{vmatrix} 1+\lambda^2 & -(1+\lambda^2) & 0 \ 1+\lambda^2 & 0 & -(1+\lambda^2) \ \lambda^2-1 & -2 & 0 \end{vmatrix} = 0$$

I know this looks intimidating, but look closer. Do you see the common factor $(1+\lambda^2)$ in the first two rows? We can pull that out!

By factoring $(1+\lambda^2)$ from the first row and $(1+\lambda^2)$ from the second, we are left with $(1+\lambda^2)^2$ multiplied by a much simpler determinant. Since $\lambda$ is a real number, $\lambda^2 \ge 0$, which means $1+\lambda^2$ is at least 1.

It is never zero! We can safely divide it away, leaving us with:

$$\begin{vmatrix} 1 & -1 & 0 \ 1 & 0 & -1 \ \lambda^2-1 & -2 & 0 \end{vmatrix} = 0$$

### The Final Resolution

Expanding this along the third column is the most efficient path. The calculation becomes a simple $2 	imes 2$ determinant:

$$0 - (-1) \begin{vmatrix} 1 & -1 \ \lambda^2-1 & -2 \end{vmatrix} + 0 = 0$$

$$1 	imes [ (1)(-2) - (-1)(\lambda^2-1) ] = 0$$

$$-2 + \lambda^2 - 1 = 0$$

$$\lambda^2 - 3 = 0$$

And there it is! The complexity collapses into the elegant result $\lambda^2 = 3$.

Solving for $\lambda$, we find $\lambda = \pm\sqrt{3}$. The set $S$ is **$\{\sqrt{3}, -\sqrt{3}\}$**. You have successfully navigated the geometry, simplified the algebra, and arrived at the truth.

Animated Solution for Mathematics - Three Dimensional Geometry: Let $S$ be the set of all real values of $λ$ such that a plane passing through the points $(- λ^{2}, 1, 1)$ , $(1, - λ^{2}, 1)$ and $(1, 1, - λ^{2})$ also passes through the point $(- 1, - 1, 1)$ . Then $S$ is equal to :

Select Answer:

Visualized Solution

Visualizing the Coplanar Points

Condition for Coplanarity

Calculating Vector $A B$

Calculating Vector $A C$

Calculating Vector $A D$

Setting up the Determinant

Simplifying the Determinant

Analyzing the Common Factor

Expanding the Determinant

Solving the 2x2 Determinant

Finding the Final Set $S$

The Sigma Insight: Equation of a Plane

Analyzing the Setup

The Power of Vectors

The Determinant Dance

The Final Resolution

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