The square of the distance of the point of intersection of the line (x-1)/2 = (y-2)/3 = (z+1)/6 and

Q: Why do we set the line equal to $\lambda$? Can't we just solve for $x$ and $y$ directly?

That is a classic point of confusion! If you try to solve for $x$ and $y$ directly, you are dealing with three variables ($x, y, z$) but only two equations from the line. It becomes a tangled mess. By introducing $\lambda$, you create a 'bridge' that allows you to express every point on the line as a function of one single variable. It turns a complex 3D geometry problem into a simple linear equation in one variable. It is the most elegant way to handle line-plane intersections.

Q: Is there a faster way to find the intersection without using the parametric form?

While you could use vector methods, the parametric form is the gold standard for JEE Advanced. It is robust, less prone to calculation errors, and works universally for any line-plane intersection. Don't look for shortcuts; master this method, and you will solve these problems with confidence and speed every single time.

Q: Why does the question ask for the square of the distance? Is it a trick?

It is not a trick, but a kindness! The standard distance formula involves a square root: $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$. By asking for $d^2$, the examiner is actually saving you from calculating a potentially messy square root. Always read the question carefully—sometimes the 'trick' is just saving you time!

Question

The square of the distance of the point of intersection of the line $\frac{x-1}{2} = \frac{y-2}{3} = \frac{z+1}{6}$ and the plane $2x - y + z = 6$ from the point $(-1, -1, 2)$ is .

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, future engineer. Today, we are not just solving a problem; we are visualizing a collision in three-dimensional space.

Imagine you are standing in a vast, empty room. In front of you, there is a flat, infinite sheet of paper—this is our plane, defined by the equation $2x - y + z = 6$.

Now, imagine a laser pointer firing a beam of light that pierces through this plane. That beam is our line, defined by $\frac{x-1}{2} = \frac{y-2}{3} = \frac{z+1}{6}$.

Our mission is to find the exact point where the laser hits the paper, and then measure how far that point is from a specific location in the room, which we call point $Q(-1, -1, 2)$.

### The Parametric Bridge

When we look at the equation of a line in symmetric form, it can feel static. It tells us the direction ratios, but it doesn't immediately give us the coordinates of a point.

To find the intersection, we need to make the line 'move.' We do this by introducing a parameter, let's call it $\lambda$. By setting the entire line equation equal to $\lambda$, we create a bridge between the abstract ratios and concrete coordinates:

$$\frac{x-1}{2} = \frac{y-2}{3} = \frac{z+1}{6} = \lambda$$

Now, watch the magic happen. We can isolate $x, y,$ and $z$ in terms of $\lambda$:

$$x = 2\lambda + 1$$
$$y = 3\lambda + 2$$
$$z = 6\lambda - 1$$

This triplet, $P(\lambda) = (2\lambda + 1, 3\lambda + 2, 6\lambda - 1)$, represents **any** point on that line. Whether $\lambda$ is $0, 1, 5,$ or $-100$, this point is guaranteed to lie on our line. We have successfully reduced a 3D line to a single variable.

### The Moment of Impact

Now, we bring the plane into the picture. The point of intersection is special because it is the only point that exists on both the line and the plane simultaneously.

If it is on the plane, it must satisfy the plane's equation: $2x - y + z = 6$. Since we know our intersection point $P$ must be of the form $(2\lambda + 1, 3\lambda + 2, 6\lambda - 1)$, we simply substitute these expressions into the plane equation:

$$2(2\lambda + 1) - (3\lambda + 2) + (6\lambda - 1) = 6$$

Take a deep breath and expand carefully:

$$4\lambda + 2 - 3\lambda - 2 + 6\lambda - 1 = 6$$

Group the $\lambda$ terms and the constants. This simplifies beautifully to:

$$7\lambda - 1 = 6$$
$$7\lambda = 7$$
$$\lambda = 1$$

We have found our parameter! The intersection occurs exactly when $\lambda = 1$.

### Locating the Point

With $\lambda = 1$ in hand, finding the coordinates of the intersection point $P$ is trivial. We plug $\lambda = 1$ back into our parametric expressions:

$$x = 2(1) + 1 = 3$$
$$y = 3(1) + 2 = 5$$
$$z = 6(1) - 1 = 5$$

So, our point of intersection is $P(3, 5, 5)$. We have successfully located the exact spot where the laser hits the plane.

### The Final Calculation

Finally, we need the square of the distance between our intersection point $P(3, 5, 5)$ and the given point $Q(-1, -1, 2)$. The distance formula in 3D is a direct extension of the Pythagorean theorem:

$$d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2$$

Substituting our coordinates:

$$d^2 = (3 - (-1))^2 + (5 - (-1))^2 + (5 - 2)^2$$
$$d^2 = (4)^2 + (6)^2 + (3)^2$$

Calculating the squares:

$$d^2 = 16 + 36 + 9$$
$$d^2 = 61$$

And there we have it. The square of the distance is **$61$**.

Animated Solution for Mathematics - Three Dimensional Geometry: The square of the distance of the point of intersection of the line $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z + 1}{6}$ and the plane $2 x - y + z = 6$ from the point $(- 1, - 1, 2)$ is .

Enter Numerical Value:

Visualized Solution

Visualizing the Setup

Parametric Form of the Line

Intersection Condition

Expanding the Equation

Solving for $λ$

Coordinates of Intersection Point

Distance Formula Setup

Substituting Coordinates

Calculating Differences

Final Computation

Conclusion

The Sigma Insight: Intersection of a Line and a Plane

Analyzing the Setup

The Parametric Bridge

The Moment of Impact

Locating the Point

The Final Calculation

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