Question

Let the line $L: \frac{x-1}{2}=\frac{y+1}{-1}=\frac{z-3}{1}$ intersect the plane $2 x+y+3 z=16$ at the point $P$ Let the point $Q$ be the foot of perpendicular from the point $R(1,-1,-3)$ on the line $L$ If $\alpha$ is the area of triangle $P Q R$, then $\alpha^2$ is equal to

Hint:

To find the area of a triangle given by three points in space, use the formula \( \frac{1}{2} \left| \mathbf{QR} \times \mathbf{QP} \right| \), where \( \mathbf{QR} \) and \( \mathbf{QP} \) are the position vectors of the sides of the triangle.

Answer 1

Correct Answer: 180

The correct answer is 180.
Any point on L((2λ+1),(−λ−1),(λ+3)) 
2(2λ+1)+(−λ−1)+3(λ+3)=16 
6λ+10=16⇒λ=1 
∴P=(3,−2,4) 
DR of QR=⟨2λ,−λ,λ+6⟩ 
DR of L=⟨2,−1,1⟩ 
4λ+λ+λ+6=0 
6λ+6=0⇒λ=−1 
Q=(−1,0,2) 

QR​×QP​=∣∣​i^24​j^​−1−2​k^−52​∣∣​=−12i^−24j^​ 
α=21​×144+576​⇒α2=4720​=180