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 $\bigtriangleup PQR$  is : Option 1) $2\sqrt{13}$ Option 2) $\frac{\sqrt{91}}{4}$ Option 3) $\frac{\sqrt{91}}{2}$ Option 4) $\frac{\sqrt{65}}{2}$

Given equation of plane is 3x-y+4z=2

Q ( 0 , -1, -3 ) is image of P

Point R is lie on plane

$DQ=\frac{|1-12-2|}{\sqrt{9+1+16}}=\frac{13}{\sqrt{26}}$

$RQ=\sqrt{(3-0)^{2}+(-1+1)^{2}+(-2+3)^{2}}=\sqrt{10}$

$\Delta QRD$ is right angled triangle

So,

$RD=\sqrt{10-\frac{13}{2}}=\sqrt{\frac{7}{2}}$

Area of $\Delta PQR=2\times (\frac{1}{2}\times \sqrt{\frac{13}{2}}\times \sqrt{\frac{7}{2}})=\frac{\sqrt{91}}{2}$

correct option (3)

Option 1)

$2\sqrt{13}$

Option 2)

$\frac{\sqrt{91}}{4}$

Option 3)

$\frac{\sqrt{91}}{2}$

Option 4)

$\frac{\sqrt{65}}{2}$

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