If the distances of the point (1, 2, a)$(1, 2, a)$ from the line (x - 1)/(1) = (y)/(2) = (z - 1)/(1)$\frac{x - 1}{1} = \frac{y}{2} = \frac{z - 1}{1}$ along the lines L₁: (x - 1)/(3) = (y - 2)/(4) = (z - a)/(b)$L_1: \frac{x - 1}{3} = \frac{y - 2}{4} = \frac{z - a}{b}$ and L₂: (x - 1)/(1) = (y - 2)/(4) = (z - a)/(c)$L_2: \frac{x - 1}{1} = \frac{y - 2}{4} = \frac{z - a}{c}$ are equal, then a + b + c$a + b + c$ is equal to

Related Formula

Core Logic

Determine the three intersection vertex positions for the matching coordinate line segments, then calculate vector cross expansions to determine face boundaries.

Step 1: Locate Intersection Vertices

Solving line pairs intersection matrices:

• L₁ L₂ A(-2, 1, 0)$L_1 \cap L_2 \implies A(-2, 1, 0)$
• L₂ L₃ B(3, 0, 1)$L_2 \cap L_3 \implies B(3, 0, 1)$
• L₃ L₁ C(0, 3, 2)$L_3 \cap L_1 \implies C(0, 3, 2)$

Step 2: Construct Vectors Cross Matrix

Using vertex values to form component arrays:

Step 3: Final Area Squared Derivation

A² = 56$A^2 = 56$

{{SOL_IMG_73}}

Pattern Recognition

Finding the area of a triangle formed by intersecting lines involves grouping directional cross vectors once coordinates are solved.

Chapter Mix

Class 12 Mathematics: Three Dimensional Geometry

Related Formula

The direction vector of a line perpendicular to two vectors u$\vec{u}$ and v$\vec{v}$ is obtained via the cross product:

Core Logic

Extract direction vectors of the given lines:

Compute the cross product:

Step 1: Write Line Equation and Intersect with Plane

Equation of line L$L$ through P(2, -1, 3)$P(2, -1, 3)$ with direction (2, -2, 1)$(2, -2, 1)$:

Any random point on this line is Q(2λ + 2, -2λ - 1, λ + 3)$Q(2\lambda + 2, -2\lambda - 1, \lambda + 3)$. For intersection with the yz$yz$-plane, set x = 0$x = 0$:

Step 2: Find Distance

Substituting λ = -1$\lambda = -1$ into the coordinate matrix of Q$Q$ gives: Q(0, 1, 2)$Q(0, 1, 2)$

Calculate distance d(P, Q)$d(P, Q)$:

Pattern Recognition

Perpendicularity to two lines always indicates using the cross-product to lock down the direction ratios. Intersection with the yz$yz$-plane simply forces x = 0$x = 0$ immediately.

Chapter Mix

Class 12 Mathematics: Three Dimensional Geometry

Related Formula

Dot product of vector projection matching orthogonal axes equals zero:

Core Logic

Let the target source coordinates tracking point match A(1, 2, 2)$A(1, 2, 2)$. General parameter points on line L$L$ are defined by parameter μ$\mu$:

Line direction vector d = i - j + 2 k$\vec{d} = \hat{i} - \hat{j} + 2\hat{k}$.

Step 1: Isolate Foot and Intersection Positions

Substituting back yields coordinate positions for foot P$P$:

Equating general vectors between standard linear constraints tracks intersection point Q$Q$ at μ = -2$\mu = -2$: Q(-1, 1, -2)$Q(-1, 1, -2)$

Step 2: Distance Formulation

Compute length of line segment squared:

Pattern Recognition

Always separate foot evaluations from line-intersection parameter updates to ensure you do not mix up variables tracking linear metrics.

Chapter Mix

Class 12 Mathematics: Three Dimensional Geometry

Related Formula

3D Cartesian distance formula:

Core Logic

Since A(x,y,z)$A(x,y,z)$ lies in the xy-plane, its z-coordinate must be zero (z = 0$z = 0$). Let the reference targets be P(0,3,2)$P(0,3,2)$, Q(2,0,3)$Q(2,0,3)$, and R(0,0,1)$R(0,0,1)$.

Setting AP² = AR²$AP^2 = AR^2$:

Step 1: Locating Coordinate Dimensions

Setting AQ² = AR²$AQ^2 = AR^2$ with y=2$y=2$:

Thus, A$A$ is precisely located at (3,2,0)$(3,2,0)$.

Step 2: Triangle Side and Area Assessment

Calculate the lengths between A(3,2,0)$A(3,2,0)$, B(1,4,-1)$B(1,4,-1)$, and C(2,0,-2)$C(2,0,-2)$: AB = √((3-1)² + (2-4)² + (0+1)²) = 3$AB = \sqrt{(3-1)^2 + (2-4)^2 + (0+1)^2} = 3$ AC = √((3-2)² + (2-0)² + (0+2)²) = 3$AC = \sqrt{(3-2)^2 + (2-0)^2 + (0+2)^2} = 3$ BC = √((1-2)² + (4-0)² + (-1+2)²) = √(18)$BC = \sqrt{(1-2)^2 + (4-0)^2 + (-1+2)^2} = \sqrt{18}$

Since AB = AC = 3$AB = AC = 3$ and AB² + AC² = BC²$AB^2 + AC^2 = BC^2$, it forms an isosceles right-angled triangle. Thus, (S1) is true.

Therefore, (S2) is false.

Pattern Recognition

Planar locations instantly zero out specific coordinate dimensions (z=0$z=0$ for xy-planes), simplifying system matrices down rapidly.

Chapter Mix

Class 11 Maths: Three Dimensional Geometry

Related Formula

Perpendicularity condition for vectors:

Core Logic

Let Q$Q$ be the projection point on the given line parameterized by λ$\lambda$: Q(2λ + 1, λ + 2, 3λ + 1)$Q(2\lambda + 1, \lambda + 2, 3\lambda + 1)$.

The vector PQ$\overrightarrow{PQ}$ from P(4,4,3)$P(4,4,3)$ is:

Step 1: Solving for Projected Intersection Points

Since PQ$\overrightarrow{PQ}$ is perpendicular to the line's direction vector (2, 1, 3)$(2, 1, 3)$:

Thus, Q$Q$ is located at (3,3,4)$(3,3,4)$.

Step 2: Transforming using Midpoint Mappings

The projection point Q$Q$ acts as the midpoint between original point P$P$ and its target image R(α, β, γ)$R(\alpha, \beta, \gamma)$:

Evaluating this gives (α, β, γ) = (2, 2, 5)$(\alpha, \beta, \gamma) = (2, 2, 5)$.

Wait, checking the options from the paper layout: option (2) represents the correct numerical matrix sum choice value 12? Let's verify the options mapping sequence matching. Ah, let's look at the calculation value carefully: 2+2+5=9$2+2+5=9$, which corresponds to choice (1).

Pattern Recognition

Midpoint properties safely speed up spatial image transitions once you locate the perpendicular projection foot.

Chapter Mix

Class 12 Maths: Three Dimensional Geometry