If `Q(0, -1, -3)` is the image of the point P in the plane `3x-y+4z=2` and R is the point (3, -1, -2), then the area (in sq units) of `DeltaPQR` is

Options

A. `(sqrt(91))/(2)`
B. `2sqrt(13)`
C. `(sqrt(91))/(4)`
D. `(sqrt(65))/(2)`

Answer

A

Explanation

Given , equation of plance `3x-y+4z=2" "(i)` and the point `Q(0,-1,-3)` is the image of point P in the plane (i), so point P is also image of point Q w.r.t. plane (i).
Let the coordinates of point P is `(x_(1), y_(1), z_(1)),` then
`(x_(1)-0)/(3)=(y_(1)+1)/(-1)=(z_(1)+3)/(4)`
`=-2([3(0)-1(-1)+4(-3)-2])/(3^(2)+(-1)^(2)+4^(2))`
"["`:.` image of the point `(x_(1), y_(1), z_(1))` in the plane `ax+by+cz+d=0` is (x, y, z), where `(x-x_(1))/(a)=(y-y_(1))/(b)=(z-z_(1))/(c)=(-2(ax_(1)+by_(1)+cz_(1)+d))/(a^(2)+b^(2)+c^(2))"]"`
`implies(x_(1)-0)/(3)=(y_(1)+1)/(-1)=(z_(1)+3)/(4)`
`=2((1-12-2))/(26)=(26)/(26)=1`
`implies" "P(x_(1), y_(1), z_(1))=(3, -2, 1)`
Now, area of `DeltaPQR`, where point `R(3, -1, -2)`
`=(1)/(2)|vec(PQ)xxvec(PR)|=(1)/(2)|"("-3hati+hatj+4hatk")"xx"("0hati+hatj-3hatk")"|`
`=(1)/(2)||{:(hati,hatj," "hatk),(-3,1,-4),(0,1,-3):}||=(1)/(2)|hati-9hatj-3hatk|`
`=(1)/(2)sqrt(1+81+9)=(sqrt(91))/(2)` sq units