The distance of the point (7, -3, -4) from the plane passing through the points (2, -3, 1), (-1, 1,

Q: Wait, why are we using a determinant? Can't I just find two vectors on the plane and take their cross product to get the normal vector?

Absolutely! That is a fantastic alternative. The cross product of $\vec{AB}$ and $\vec{AC}$ gives you the normal vector $\vec{n} = (a, b, c)$. The determinant method is essentially a compact way of writing the scalar triple product equation $(\vec{r} - \vec{A}) \cdot (\vec{AB} \times \vec{AC}) = 0$. Both paths lead to the same destination—choose the one that feels more intuitive to you!

Q: Does it matter which point I choose as $(x_1, y_1, z_1)$ in the determinant?

Not at all! The plane is a unique geometric object. Whether you pick point $A$, $B$, or $C$ as your reference point, the resulting equation will be equivalent (perhaps just scaled by a constant factor). The plane's orientation in space remains unchanged.

Q: I'm confused about the distance formula. Why do we divide by the square root of $a^2 + b^2 + c^2$?

Think of the plane equation $ax + by + cz + d = 0$ as defining a normal vector $\vec{n} = (a, b, c)$. When we project the vector connecting a point on the plane to our target point $P$ onto this normal vector, we are essentially normalizing the projection. That square root term is the magnitude of the normal vector, which scales our result to give the true perpendicular distance.

Question

The distance of the point $(7, -3, -4)$ from the plane passing through the points $(2, -3, 1)$, $(-1, 1, -2)$ and $(3, -4, 2)$ is :

Options:

(A) 4

(B) 5

(C) $5\sqrt{2}$

(D) $4\sqrt{2}$

Sigma Percentile · Accepted Answer

### The Geometry of Space: Setting the Stage

Imagine you are floating in a vast, three-dimensional void. Before you, there is a flat, infinite sheet of paper—a plane. This plane is anchored by three specific points in space: $A(2, -3, 1)$, $B(-1, 1, -2)$, and $C(3, -4, 2)$.

Hovering above this plane is a target point, $P(7, -3, -4)$. Our mission is to find the shortest possible distance from $P$ to the plane.

In the world of JEE Advanced, this is a classic problem of spatial intuition. We are not just calculating numbers; we are defining the relationship between a point and a surface.

### The Mathematical Address: Finding the Plane

To measure the distance, we first need to know exactly where our plane is. We need its 'mathematical address'—the Cartesian equation $ax + by + cz + d = 0$.

Since we have three points, we can use the elegant determinant method. By setting the determinant of the vectors formed by these points to zero, we lock the plane in place:

$$
\begin{vmatrix} x-x_1 & y-y_1 & z-z_1 \ x_2-x_1 & y_2-y_1 & z_2-z_1 \ x_3-x_1 & y_3-y_1 & z_3-z_1 \end{vmatrix} = 0
$$

Let's substitute our points $A$, $B$, and $C$ into this structure. Using $A(2, -3, 1)$ as our reference point, we get:

$$
\begin{vmatrix} x-2 & y+3 & z-1 \ -3 & 4 & -3 \ 1 & -1 & 1 \end{vmatrix} = 0
$$

Notice how the arithmetic simplifies beautifully. The second row becomes $(-3, 4, -3)$ and the third row becomes $(1, -1, 1)$. This is the moment where the complexity starts to melt away.

### The Vanishing Act: Simplifying the Equation

Now, let's expand this determinant along the first row. For the $(x-2)$ term, we have $(4)(1) - (-1)(-3) = 4 - 3 = 1$.

For the $(y+3)$ term, we have $(-3)(1) - (1)(-3) = -3 + 3 = 0$. For the $(z-1)$ term, we have $(-3)(-1) - (1)(4) = 3 - 4 = -1$.

Look at that! The middle term, the one with $y$, completely vanishes because $-3 + 3 = 0$. This leaves us with a beautifully simple equation:

$$
(x-2)(1) - (y+3)(0) + (z-1)(-1) = 0
$$

This simplifies to $x - 2 - z + 1 = 0$, or simply **$x - z - 1 = 0$**. This is the exact equation of our plane.

### The Final Drop: Calculating the Distance

Now that we have our plane $x - z - 1 = 0$, we return to our target point $P(7, -3, -4)$. The perpendicular distance $d$ from a point $(x_0, y_0, z_0)$ to a plane $ax + by + cz + d = 0$ is given by the formula:

$$
d = \frac{|ax_0 + by_0 + cz_0 + d|}{\sqrt{a^2 + b^2 + c^2}}
$$

Plugging in our values, where $a=1, b=0, c=-1, d=-1$, and our point $P(7, -3, -4)$, we get:

$$
d = \frac{|(1)(7) + (0)(-3) + (-1)(-4) - 1|}{\sqrt{(1)^2 + (0)^2 + (-1)^2}}
$$

The numerator becomes $|7 + 0 + 4 - 1| = |10| = 10$. The denominator is $\sqrt{1 + 0 + 1} = \sqrt{2}$.

Thus, $d = \frac{10}{\sqrt{2}}$. To finish, we rationalize the denominator:

$$
d = \frac{10\sqrt{2}}{2} = 5\sqrt{2}
$$

And there it is! The shortest distance from our point to the plane is **$5\sqrt{2}$ units**. It is a perfect example of how, with a clear geometric strategy and careful algebraic execution, even the most daunting 3D problems collapse into elegant, simple solutions.

Animated Solution for Mathematics - Three Dimensional Geometry: The distance of the point $(7, - 3, - 4)$ from the plane passing through the points $(2, - 3, 1)$ , $(- 1, 1, - 2)$ and $(3, - 4, 2)$ is :

Select Answer:

Visualized Solution

The Geometric Setup

Equation of a Plane through 3 Points

Substituting the Coordinates

Simplifying the Rows

Expanding the Determinant

The Final Equation of the Plane

The Perpendicular Distance Formula

Substituting into the Distance Formula

Calculating the Result

Final Answer

The Sigma Insight: Equation of a Plane

The Geometry of Space

The Mathematical Address

The Vanishing Act

The Final Drop

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