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Let the foot of perpendicular of the point \mathrm{P}(3,-2,-9) on the plane passing through the points (-1,-2,-3),(9,3,4),(9,-2,1)be Q(\alpha, \beta, \gamma). Then the distance of Q from the origin is

Option: 1

\sqrt{29}


Option: 2

\sqrt{38}


Option: 3

\sqrt{42}


Option: 4

\sqrt{35}


Answers (1)

best_answer

Equation of plane through \mathrm{A}, \mathrm{B}, \mathrm{C}.
\left|\begin{array}{ccc} x+1 & y+2 & z+3 \\ 10 & 5 & 7 \\ 10 & 0 & 4 \end{array}\right|=0
2 x+3 y-5 z-7=0
Foot of \mathrm{I}^{\mathrm{r}}$ of $\mathrm{P}(3,-2,-9) is
\begin{aligned} & \frac{x-3}{2}=\frac{y+2}{3}=\frac{z+9}{-5}=-\frac{(6-\phi+45-7)}{4+9+25} \\ & \frac{x-3}{2}=\frac{y+2}{3}=\frac{z+9}{-5}=-1 \\ & \mathrm{Q}(1,-5,-4)=(\alpha, \beta, \gamma) \\ & \mathrm{OQ}=\sqrt{\alpha^2+\beta^2+\gamma^2}=\sqrt{42} \end{aligned}

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Suraj Bhandari

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