Let Q and R be two points on the line ( x +1/2)=( y +2/3)=( z -1/2) at a distance √26 from the point P (4,2,7). Then the square of the area of the triangle PQR is .

Question

Let Q and R be two points on the line ( x +1/2)=( y +2/3)=( z -1/2) at a distance √26 from the point P (4,2,7). Then the square of the area of the triangle PQR is . (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

Let (2 λ-1,3 λ-2,2 λ+1) be any point on the line (2 λ-5)2+(3 λ-4)2+(2 λ-6)2=26 λ=1,3 Q (1,1,3) ; R (5,7,7) ; P (4,2,7) Area of triangle PQR =1 / 2| PQ × PR | =√153

Q. 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 $26$ from the point $P (4, 2, 7)$ . Then the square of the area of the triangle $PQR$ is _______.

Let $(2 λ - 1, 3 λ - 2, 2 λ + 1)$ be any point on the line
$(2 λ - 5)^{2} + (3 λ - 4)^{2} + (2 λ - 6)^{2} = 26$
$λ = 1, 3$
$Q (1, 1, 3); R (5, 7, 7); P (4, 2, 7)$
Area of triangle $PQR = 1/2∣ PQ \times PR ∣$
$= 153$

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

Let (2λ−1,3λ−2,2λ+1) be any point on the line (2λ−5)2+(3λ−4)2+(2λ−6)2=26 λ=1,3 Q(1,1,3);R(5,7,7);P(4,2,7) Area of triangle PQR=1/2∣PQ​×PR∣ =153​

Q. 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 $26$ from the point $P (4, 2, 7)$ . Then the square of the area of the triangle $PQR$ is _______.

Let $(2 λ - 1, 3 λ - 2, 2 λ + 1)$ be any point on the line
$(2 λ - 5)^{2} + (3 λ - 4)^{2} + (2 λ - 6)^{2} = 26$
$λ = 1, 3$
$Q (1, 1, 3); R (5, 7, 7); P (4, 2, 7)$
Area of triangle $PQR = 1/2∣ PQ \times PR ∣$
$= 153$