Q.A rectangular parallelepiped (box) is drawn with its edges parallel to the coordinate axes, one vertex at the origin $O(0,0,0)$ and the diagonally opposite vertex at $P(2,4,5)$; the three edges from $O$ run along the $x$-, $y$- and $z$-axes. Let $F$ be the vertex of this box that is the foot of the perpendicular dropped from $P$ onto the XZ-plane (the plane $y=0$) — i.e. the corner having the same $x$- and $z$-coordinates as $P$ but lying in the plane $y=0$. Find the coordinates of $F$.

Concept & Intuition

The YZ-plane is the set of all points where $x = 0$. The perpendicular from any point to this plane is simply the line parallel to the x-axis, so the foot $L$ is obtained by dropping the x-coordinate to zero while keeping the y- and z-coordinates unchanged. Once we have $L$, its distance from the origin is just the square root of the sum of squares of its y- and z-coordinates. Setting that equal to $\sqrt{41}$ gives a quadratic in $k$.

Step-by-step solution

1. Find the foot $L$ of the perpendicular from $P$ to the YZ-plane. The YZ-plane is defined by $x = 0$. The perpendicular from any point to this plane is horizontal (parallel to the x-axis), so the foot has the same y- and z-coordinates as $P$, but x-coordinate 0.

2. Write the distance from $L$ to the origin $O(0,0,0)$. The distance formula in 3D gives

3. Set this distance equal to $\sqrt{41}$ and solve for $k$.

Squaring both sides (both sides are non-negative, so no extraneous solutions are introduced):

4. Solve the quadratic equation. Taking square roots:

• Case 1: $3k-7 = 5 \;\Rightarrow\; 3k = 12 \;\Rightarrow\; k = 4$.
• Case 2: $3k-7 = -5 \;\Rightarrow\; 3k = 2 \;\Rightarrow\; k = \frac{2}{3}$.

5. Check which value appears among the options. The options are: (A) 4, (B) 1, (C) $-\frac{2}{3}$, (D) $\frac{11}{3}$. We have $k = 4$ and $k = \frac{2}{3}$. Only $k = 4$ is listed.

Concept and Intuition

To find the image of a point in a line, we treat the line as a mirror. The image is the point such that the line is the perpendicular bisector of the segment joining the point and its image.

So the key steps are:

• Find the foot of the perpendicular from the point to the line.
• Use the foot as the midpoint between the point and its image.

This works because reflection in a line is equivalent to a 180° rotation about that line — the line stays fixed, and the point’s perpendicular distance to the line is preserved.

Step-by-step solution

1. Parameterize the line

The line is given by

So any point on the line is

2. Find the foot of the perpendicular from $P(3,5,3)$ to the line

Let the foot be $F(t)$. The vector from $P$ to $F$ is

The direction vector of the line is $\vec{d} = (1,2,3)$.

For $F$ to be the foot, $\overrightarrow{PF}$ must be perpendicular to $\vec{d}$, so their dot product is zero:

Thus the foot is

3. Use the midpoint property

The foot $F$ is the midpoint of $P$ and its image $P'(a,b,c)$:

Solving:

So $P' = (-1,1,7)$.

4. Compute $a+b+c$

Concept: family of planes through the line of intersection of P1 = 0 and P2 = 0 is P1 + k*P2 = 0; fix k using the given distance condition.

Plane: (x + 2y + 3z - 2) + k(x - y + z - 3) = 0

=> (1 + k)x + (2 - k)y + (3 + k)z - (2 + 3k) = 0.

Distance from (3, 1, -1):

Numerator = |3(1 + k) + 1(2 - k) + (-1)(3 + k) - (2 + 3k)|

= |3 + 3k + 2 - k - 3 - k - 2 - 3k| = |-2k| = 2|k|.

Denominator = sqrt((1 + k)^2 + (2 - k)^2 + (3 + k)^2)

= sqrt(1 + 2k + k^2 + 4 - 4k + k^2 + 9 + 6k + k^2)

= sqrt(3k^2 + 4k + 14).

Set 2|k| / sqrt(3k^2 + 4k + 14) = 2/sqrt(3).

Square: 4k^2 / (3k^2 + 4k + 14) = 4/3

=> 12k^2 = 12k^2 + 16k + 56

=> 16k = -56 => k = -7/2.

Substitute k = -7/2:

(1 - 7/2)x + (2 + 7/2)y + (3 - 7/2)z - (2 - 21/2) = 0

=> (-5/2)x + (11/2)y - (1/2)z + 17/2 = 0.

Multiply by -2: 5x - 11y + z - 17 = 0, i.e. 5x - 11y + z = 17.

ANSWER: A

Concept: parametrise both lines and solve for the parameters that make the points coincide.

Line 1: (x - 1)/2 = (y - 2)/3 = (z - 3)/4 = t

=> (1 + 2t, 2 + 3t, 3 + 4t).

Line 2: (x - 4)/5 = (y - 1)/2 = z/1 = s

=> (4 + 5s, 1 + 2s, s).

Equate the z-coordinates: 3 + 4t = s.

Equate the y-coordinates: 2 + 3t = 1 + 2s => 3t - 2s = -1.

Substitute s = 3 + 4t: 3t - 2(3 + 4t) = -1 => 3t - 6 - 8t = -1 => -5t = 5 => t = -1, and s = 3 + 4(-1) = -1.

Check the x-coordinates: 1 + 2(-1) = -1 and 4 + 5(-1) = -1. Consistent.

Point of intersection: (1 - 2, 2 - 3, 3 - 4) = (-1, -1, -1).

ANSWER: C

Take the pencil of planes through the intersection:

Substitute $(1,0,1)$: $(2+0+1-7)+\lambda(1+0+4+5)=-4+10\lambda=0\Rightarrow\lambda=\tfrac25$.

Then $5(2x+y+z-7)+2(x+3y+4z+5)=0$:

This is the well-known problem with plane $3x-6y+2z+11=0$ (normal $\hat n=3\hat i-6\hat j+2\hat k$, $|\hat n|=\sqrt{9+36+4}=7$). The distance of $(2,3,4)$ is