# 14 In the following cases, find the distance of each of the given points from the corresponding given plane POINT PLANE a. (0, 0, 0) 3x – 4y + 12 z = 3 b. (3, – 2, 1) 2x – y + 2z + 3 = 0 c. (2, 3, – 5) x + 2y – 2z = 9 d. (– 6, 0, 0) 2x – 3y + 6z – 2 = 0

We know that the distance between a point $P (x_{1},y_{1},z_{1})$ and a plane $Ax+By+Cz =D$ is given by,

$d =\left | \frac{Ax_{1}+By_{1}+Cz_{1}-D}{\sqrt{A^2+B^2+C^2}} \right |$                                   .......................(1)

So, calculating for each case;

(a) Point $(0,0,0)$ and Plane $3x-4y+12z = 3$

Therefore, $d =\left | \frac{3(0)-4(0)+12(0)-3}{\sqrt{3^2+(-4)^2+12^2}} \right | = \frac{3}{\sqrt{169}} = \frac{3}{13}$

(b) Point $(3,-2,1)$ and Plane $2x-y+2z +3= 0$

Therefore, $d =\left | \frac{2(3)-(-2)+2(1)+3}{\sqrt{2^2+(-1)^2+2^2}} \right | = \frac{13}{3}$

(c) Point $(2,3,-5)$ and Plane $x+2y-2z =9$

Therefore, $d =\left | \frac{2+2(3)-2(-5)-9}{\sqrt{1^2+2^2+(-2)^2}} \right | = \frac{9}{3} = 3$

(d) Point $(-6,0,0)$ and Plane $2x-3y+6z -2= 0$

Therefore, $d =\left | \frac{2(-6)-3(0)+6(0)-2}{\sqrt{2^2+(-3)^2+6^2}} \right | = \frac{-14}{\sqrt{49}} = \frac{14}{7} =2$

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