If the equation of the plane passing through the line of intersection of the planes 2x - y + z = 3,

Q: Wait, why do we use the $P_1 + \lambda P_2 = 0$ form? Can't we just find the line of intersection first?

You absolutely could find the line of intersection first, but that would involve finding the direction vector of the line (via the cross product of the two plane normals) and finding a specific point on the line. That is a much longer, more tedious path. The family of planes method is the 'elegant' way—it captures all possible planes passing through that line in one single, powerful equation. It is like using a master key instead of picking the lock.

Q: Why is the dot product of the normal vector and the line's direction vector zero? Shouldn't they be parallel?

That is a classic trap! Remember, the normal vector $\vec{n}$ is perpendicular to the plane itself. If the plane is parallel to the line, the line is 'lying' within the plane's orientation. Therefore, the normal vector (which sticks out of the plane at 90 degrees) must be perpendicular to the line. So, $\vec{n} \cdot \vec{d} = 0$ is the correct condition.

Q: Why did we have to divide the final equation by 2? Does that change the plane?

Great observation! Dividing the equation by a constant does not change the plane itself—it just scales the coefficients. The plane $22x - 14y + 20z + 12 = 0$ is geometrically identical to $11x - 7y + 10z + 6 = 0$. We divide by 2 simply to match the specific format $ax + by + cz + 6 = 0$ requested by the problem. Always keep an eye on the target format!

Question

If the equation of the plane passing through the line of intersection of the planes $2x - y + z = 3, 4x - 3y + 5z + 9 = 0$ and parallel to the line $\frac{x+1}{-2} = \frac{y+3}{4} = \frac{z-2}{5}$ is $ax + by + cz + 6 = 0$, then $a + b + c$ is equal to

Options:

(A) 12

(B) 14

(C) 16

(D) 13

Sigma Percentile · Accepted Answer

### The Architecture of 3D Space: Mastering the Family of Planes

Welcome, fellow traveler in the realm of JEE Advanced mathematics. Today, we are not just solving an equation; we are performing an act of architectural design in three-dimensional space.

We are given two planes, $P_1: 2x - y + z - 3 = 0$ and $P_2: 4x - 3y + 5z + 9 = 0$. Imagine these as two massive, infinite sheets of paper slicing through the void. Where they meet, they create a line—a sharp, definitive edge.

Our mission is to find a third plane that passes through this exact edge, but with a specific orientation: it must be parallel to a given line $L$. This is a problem of constraints, and it is a beautiful one.

### Phase 1: The Hinge of Infinity

When we talk about a plane passing through the intersection of two others, we are essentially talking about a door swinging on a hinge. The line of intersection is our hinge, and there are infinitely many planes that can rotate around it.

We capture this infinite family using the equation $P_1 + \lambda P_2 = 0$. By substituting our given planes, we obtain:

$$(2x - y + z - 3) + \lambda(4x - 3y + 5z + 9) = 0$$

Think of $\lambda$ as the "tuning knob." As you change the value of $\lambda$, the plane rotates around that intersection line. Our goal is to find the specific value of $\lambda$ that locks this plane into the perfect position—the one where it runs parallel to our target line.

### Phase 2: The DNA of the Plane

To control the orientation of our plane, we must look at its "DNA"—its normal vector, $\vec{n}$. The normal vector is the vector that points straight out of the plane, defining its tilt and direction.

Let us group our equation by the variables $x$, $y$, and $z$:

$$(2 + 4\lambda)x + (-1 - 3\lambda)y + (1 + 5\lambda)z + (9\lambda - 3) = 0$$

From this, we extract the normal vector: $\vec{n} = \langle 2 + 4\lambda, -1 - 3\lambda, 1 + 5\lambda \rangle$. This vector is the key to our puzzle, as it dictates exactly how our plane is oriented in space.

### Phase 3: The Parallelism Constraint

Now, we bring in the external constraint. We are told our plane must be parallel to the line $L$ defined by:

$$\frac{x+1}{-2} = \frac{y+3}{4} = \frac{z-2}{5}$$

The direction vector of this line is $\vec{d} = \langle -2, 4, 5 \rangle$.

If a plane is parallel to a line, the line does not pierce the plane; it glides alongside it. This means the normal vector of the plane must be perpendicular to the direction vector of the line. In the language of vectors, their dot product must vanish: $\vec{n} \cdot \vec{d} = 0$.

Performing this operation:

$$-2(2 + 4\lambda) + 4(-1 - 3\lambda) + 5(1 + 5\lambda) = 0$$

### Phase 4: The Resolution

Expanding the terms systematically:

$$-4 - 8\lambda - 4 - 12\lambda + 5 + 25\lambda = 0$$

Combining the constants and the $\lambda$ terms:

$$-3 + 5\lambda = 0 \Rightarrow \lambda = \frac{3}{5}$$

We have found the value that locks our door in place.

### Phase 5: The Final Polish

Substituting $\lambda = \frac{3}{5}$ back into our grouped equation:

$$(2 + 4(\frac{3}{5}))x + (-1 - 3(\frac{3}{5}))y + (1 + 5(\frac{3}{5}))z + (9(\frac{3}{5}) - 3) = 0$$

Simplifying this yields:

$$22x - 14y + 20z + 12 = 0$$

The problem asks for the form $ax + by + cz + 6 = 0$. Dividing the entire equation by $2$, we get:

$$11x - 7y + 10z + 6 = 0$$

Comparing this to the required form, we identify $a = 11$, $b = -7$, and $c = 10$. The final sum, $a + b + c$, is:

$$11 - 7 + 10 = \mathbf{14}$$

Select Answer:

Visualized Solution

Visualizing the Intersecting Planes

The Family of Planes

Substituting the Plane Equations

Extracting the Normal Vector $n$

The Parallel Line Condition

Applying the Dot Product $n \cdot d = 0$

Setting Up the Equation for $λ$

Solving for $λ$

Substituting $λ$ Back

Simplifying the Plane Equation

Normalizing to the Target Form

Final Calculation of $a + b + c$

The Sigma Insight: Equation of a Plane

The Architecture of 3D Space

Phase 1

Phase 2

Phase 3

Phase 4

Phase 5

Similar Questions

Let the plane passing through the point $(- 1, 0, - 2)$ and perpendicular to each of the planes $2 x + y - z = 2$ and $x - y - z = 3$ be $a x + b y + cz + 8 = 0$ . Then the value of $a + b + c$ is equal to:

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The equation of the plane containing the straight line $\frac{x}{2} = \frac{y}{3} = \frac{z}{4}$ and perpendicular to the plane containing the straight lines $\frac{x}{3} = \frac{y}{4} = \frac{z}{2}$ and $\frac{x}{4} = \frac{y}{2} = \frac{z}{3}$ is:

If the equation of a plane $P$ , passing through the intersection of the planes $x + 4 y - z + 7 = 0$ and $3 x + y + 5 z = 8$ is $a x + b y + 6 z = 15$ for some $a, b \in R$ , then the distance of the point $(3, 2, - 1)$ from the plane $P$ is

The vector equation of the plane through the line of intersection of the planes $x + y + z = 1$ and $2 x + 3 y + 4 z = 5$ which is perpendicular to the plane $x - y + z = 0$ is :

List-I

List-II