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Step-by-Step Solution
Line equation:
The given line is
L:\frac{x + 1}{2} = \frac{y + 2}{3} = \frac{z - 1}{2}.
A convenient parametric form of this line can be written as
x = 2t - 1,\quad y = 3t - 2,\quad z = 2t + 1.
Step 1: Express a general point on the line
Let T(t) = (2t - 1,\, 3t - 2,\, 2t + 1) be a general point on the line.
Step 2: Impose the distance condition
We know Q and R lie on this line at a distance \sqrt{26} from
P(4, 2, 7) . Hence, the distance PT must satisfy
PT = \sqrt{(2t - 1 - 4)^2 + (3t - 2 - 2)^2 + (2t + 1 - 7)^2}
= \sqrt{26}.
This leads to the equation
(2t - 5)^2 + (3t - 4)^2 + (2t - 6)^2 = 26.
Step 3: Solve for the parameter t
Expand and simplify:
\[
(2t - 5)^2 = 4t^2 - 20t + 25,
\]
\[
(3t - 4)^2 = 9t^2 - 24t + 16,
\]
\[
(2t - 6)^2 = 4t^2 - 24t + 36.
\]
Adding these up:
\[
4t^2 + 9t^2 + 4t^2 = 17t^2,\quad
(-20t) + (-24t) + (-24t) = -68t,\quad
25 + 16 + 36 = 77.
\]
So
\[
17t^2 - 68t + 77 = 26
\quad \Longrightarrow \quad
17t^2 - 68t + 51 = 0.
\]
Factoring,
\[
17(t^2 - 4t + 3) = 0
\quad\Longrightarrow\quad
t^2 - 4t + 3 = 0
\quad\Longrightarrow\quad
(t - 3)(t - 1) = 0.
\]
Hence, t = 1 or t = 3.
Step 4: Identify points Q and R
Corresponding to t=1 :
\[
Q = (2(1)-1,\,3(1)-2,\,2(1)+1) = (1,\,1,\,3).
\]
Corresponding to t=3 :
\[
R = (2(3)-1,\,3(3)-2,\,2(3)+1) = (5,\,7,\,7).
\]
Thus, points Q(1,1,3) and R(5,7,7) both lie on the line and are
at distance \sqrt{26} from P(4,2,7) .
Step 5: Compute vectors and use the formula for the area of a triangle in 3D
Let PQ = Q - P and PR = R - P :
\[
PQ = (1 - 4,\ 1 - 2,\ 3 - 7) = (-3,\ -1,\ -4),
\]
\[
PR = (5 - 4,\ 7 - 2,\ 7 - 7) = (1,\ 5,\ 0).
\]
The area of the triangle PQR is
\[
\text{Area} = \frac{1}{2} \bigl\lVert \overrightarrow{PQ} \times \overrightarrow{PR} \bigr\rVert.
\]
Step 6: Calculate the cross product PQ \times PR
\[
\overrightarrow{PQ} \times \overrightarrow{PR}
=
\begin{vmatrix}
\mathbf{i} & \mathbf{j} & \mathbf{k} \\
-3 & -1 & -4 \\
1 & 5 & 0
\end{vmatrix}.
\]
Expanding,
\[
= \mathbf{i}\,\bigl((-1)\cdot 0 - (-4)\cdot 5\bigr)
- \mathbf{j}\,\bigl((-3)\cdot 0 - (-4)\cdot 1\bigr)
+ \mathbf{k}\,\bigl((-3)\cdot 5 - (-1)\cdot 1\bigr).
\]
\[
= \mathbf{i}\,(0 + 20) - \mathbf{j}\,(0 + 4) + \mathbf{k}\,(-15 + 1).
\]
\[
= (20,\ -4,\ -14).
\]
So,
\[
\lVert \overrightarrow{PQ} \times \overrightarrow{PR} \rVert
= \sqrt{20^2 + (-4)^2 + (-14)^2}
= \sqrt{400 + 16 + 196}
= \sqrt{612}
= \sqrt{4 \times 153}
= 2\,\sqrt{153}.
\]
Step 7: Find the area and then its square
Hence, the area of \triangle PQR is
\[
\text{Area} = \frac{1}{2} \times 2\sqrt{153} = \sqrt{153}.
\]
Therefore, the square of the area is
\[
(\sqrt{153})^2 = 153.
\]
Hence, the required square of the area of the triangle is 153.
