If two distinct point $\mathrm{Q}, \mathrm{R}$ lie on the line of intersection of the planes $-x+2 y-z=0$ and $3 x-5 y+2 z=0$ and $P Q=P R=\sqrt{18}$ where the point P is $(1,-2,3)$, then the area of the triangle PQR is equal to

JEE Main · 2022 · 28 Jun 2022 · Shift 1 · Three Dimensional Geometry

If two distinct point

Q, R

lie on the line of intersection of the planes

- x + 2 y - z = 0

and

3 x - 5 y + 2 z = 0

and

P Q = P R = 18

where the point P is

(1, - 2, 3)

, then the area of the triangle PQR is equal to

Answer: (B)

The direction vector of the line of intersection is

n = \hat{i} - 1 3 \hat{j} 2 - 5 \hat{k} - 1 2 = - \hat{i} - \hat{j} - \hat{k}

, so the line is

\frac{x}{1} = \frac{y}{1} = \frac{z}{1}

.

Let

T

be the foot of perpendicular from

P (1, - 2, 3)

to the line. Using the perpendicularity condition, we find

T (\frac{2}{3}, \frac{2}{3}, \frac{2}{3})

and

P T = \frac{114}{3}

.

Since

P Q = P R = 18

and

T

is the midpoint of

QR

, we have an isosceles triangle. Using

cos θ = \frac{P T}{P Q} = \frac{19}{3 3}

, we find

sin 2 θ = \frac{4 38}{27}

.

Area

= \frac{1}{2} \times P Q \times P R \times sin 2 θ = \frac{1}{2} \times 18 \times \frac{4 38}{27} = \frac{4}{3} 38

JEE Main · 2022 · 28 Jun 2022 · Shift 1 · Single correct option