Question

Let three points $P(3,2,-1),Q(1,3,4)$ and $R(2,1,-2)$ and the distance of point $P$ from the plane $OQR$ is $d.$ Then $3d+3=$
$($Where $O$ is the origin$)$

• A. $7$
• B. $6$
• C. $5$
• D. $4$

Answer 1

The correct option is D $4$

Given points: $O(0,0,0),P(3,2,-1),Q(1,3,4)$ and $R(2,1,-2)$
equation of plane $OQR$ given by,
$\left|\begin{array}{ccc}x-x_1& y-y_1& z-z_1\\ x_2-x_1& y_2-y_1& z_2-z_1\\ x_3-x_1& y_3-y_1& z_3-z_1\end{array}\right|=0\Rightarrow \left|\begin{array}{ccc}x-0& y-0& z-0\\ 1-0& 3-0& 4-0\\ 2-0& 1-0& -2-0\end{array}\right|=0\Rightarrow 2x-2y+z=0$
so, dostance of point $P$ from the above plane is :
$d=\frac{|2\left(3\right)-2\left(2\right)+1(-1)|}{\sqrt{2^2+2^2+1^2}}\Rightarrow d=\frac{1}{3}\therefore 3d+3=4$