Three Dimensional Geometry — Class 12 Maths Solution

.: (a) The given plane is $3x + 4y + 12z = 3$...(1)

$\therefore$ Distance of $(0,0,0)$ from plane (1)

$= \left| {\cfrac{{3(0) - 4(0) + 12(0) - 3}}{{\sqrt {9 + 16 + 144} }}} \right| = \left| {\cfrac{{ - 3}}{{\sqrt {169} }}} \right| = \cfrac{3}{{13}}$

units

(b) The given plane is $2x - y + 2z + 3 = 0$

Distance of $(3, - 2,1)$

from plane (1)
$= \left| {\cfrac{{2(3) - ( - 2) + 2(1) + 3}}{{\sqrt {4 + 1 + 4} }}} \right| = \left| {\cfrac{{6 + 2 + 2 + 3}}{{\sqrt 9 }}} \right| = \cfrac{{13}}{3}$ units

(c) The given plane is $x + 2y - 2z - 9 = 0$ …(1)

Distance of $(2,3, - 5)$ from plane (1)

$= \left| {\cfrac{{2 + 2(3) - 2( - 5) + ( - 9)}}{{\sqrt {1 + 4 + 4} }}} \right|$

$= \left| {\cfrac{{2 + 6 + 10 - 9}}{{\sqrt 9 }}} \right| = \left| {\cfrac{9}{3}} \right| = 3$ units

(d) The given plane is $2x - 3y + 6z - 2 = 0$

$\therefore$ Distance of $( - 6,0,0)$ from plane (1)

$= \left| {\cfrac{{2( - 6) - 3(0) + 6(0) - 2}}{{\sqrt {4 + 9 + 36} }}} \right| = \left| {\cfrac{{ - 12 - 0 + 0 - 2}}{{\sqrt {49} }}} \right| = \cfrac{{14}}{7} = 2$ units

Miscellaneous Exercise