Question

Let Q and R be two points on the line
$\frac{x+1}{2}=\frac{y+2}{3}=\frac{z−1}{2}$ at a distance $\sqrt{26}$
from the point P(4, 2, 7). Then the square of the area of the triangle PQR is _______ .

Answer 1

The correct answer is 153
$L: \frac{x+1}{2}=\frac{y+2}{3}=\frac{z−1}{2}$

Fig

Let T(2 t – 1, 3 t – 2, 2 t + 1)
$∵ PT⊥QR$
∴ 2(2 t – 5) + 3 (3 t – 4) + 2 (2 t – 6) = 0
17 t = 34
t = 2
So T(3, 4, 5)
$∴ PT=\sqrt{1+4+4}=3$
$∴ QT=\sqrt{26−9}=\sqrt{17}$
∴Area of $△PQR=\frac{1}{2}×2\sqrt{17}×3=3\sqrt{17}$
∴Square of $ar(△PQR)=153.$