Question

The points $A( - 1,3,0)$, $B(2,2,1)$ and $C(1,1,3)$ determine a plane. The distance from the plane to the point $D(5,7,8)$ is

• A. $\sqrt{66}$
• B. $\sqrt{71}$
• C. $\sqrt{73}$
• D. $\sqrt{76}$

Answer 1

Answer: (A)

$\sqrt{66}$

Answer 2

Equation of plane passing through $( - 1,3,0 )$ is

$A ( x + 1 ) + B ( y - 3 ) + C ( z - 0 ) = 0$ ......(i)

Also, plane (i) is passing through the points $( 2,2,1 )$ and
(1, 1, 3). So, $3 A - B + C = 0$ .....(ii)

$2 A - 2 B + 3 C = 0$ .....(iii)

Solving (ii) and (iii), $\frac { A } { - 3 + 2 } = \frac { B } { 2 - 9 } = \frac { C } { - 6 + 2 }$

∴ $A : B : C = - 1 : - 7 : - 4$ or $A : B : C = 1 : 7 : 4$

From (i), $1 ( x + 1 ) + 7 ( y - 3 ) + 4 ( z ) = 0$ or $x + 7 y + 4 z - 20 = 0$

Distance from the plane to the point (5, 7, 8) is, $\frac { 1 \times 5 + 7 \times 7 + 4 \times 8 - 20 } { \sqrt { 1 ^ { 2 } + 7 ^ { 2 } + 4 ^ { 2 } } } = \frac { 5 + 49 + 32 - 20 } { \sqrt { 66 } } = \frac { 66 } { \sqrt { 66 } } = \sqrt { 66 }$