3D Geometry

Area of $\Delta $ABC = ${1 \over 2} \times BC \times AD$

Given BC = 5 so we need perpendicular distance of A from line BC.

Let a point D on BC = (3$\lambda $ - 2, 1, 4$\lambda $)

Direction ratio of AD

3$\lambda $ - 3, 2, 4$\lambda $ - 2

As AD perpendicular to the BC, so

DR of AD. DR of BC = 0

$(3\lambda - 3)3 + 2(0) + (4\lambda - 2)4 = 0$

$9\lambda - 9 + 16\lambda - 8 = 0 \Rightarrow \lambda = {{17} \over {25}}$

Hence, $D = \left( {{1 \over {25}},1,{{68} \over {25}}} \right)$

$\left| {\overline {AD} } \right| = \sqrt {{{\left( {{1 \over {25}} - 1} \right)}^2} + {{\left( 2 \right)}^2} + {{\left( {{{68} \over {25}} - 2} \right)}^2}} $

$ \Rightarrow \sqrt {{{\left( {{{ - 24} \over {25}}} \right)}^2} + 4 + {{\left( {{{18} \over {25}}} \right)}^2}} $

$ \Rightarrow \sqrt {{{{{\left( {24} \right)}^2} + 4{{\left( {25} \right)}^2} + {{\left( {18} \right)}^2}} \over {{{25}^2}}}} $

$ \Rightarrow \sqrt {{{576 + 2500 + 324} \over {{{25}^2}}}} $

$ \Rightarrow \sqrt {{{3400} \over {{{25}^2}}}} \Rightarrow {{\sqrt {34} .10} \over {25}} = {{2\sqrt {34} } \over 5}$

Area of triangle = ${1 \over 2} \times \left| {\overline {BC} } \right| \times \left| {\overline {AD} } \right|$

= ${1 \over 2} \times 5 \times {{2\sqrt {34} } \over 5} = \sqrt {34} $

P₁ : x + y + z – 6 = 0

P₂ : 2x + 3y + z + 5 = 0

Equation of plane which passes through the line of intersection of P₁ and P₂ is

P₁ + $\lambda $P₂ = 0

$ \Rightarrow $ (x + y + z – 6) + $\lambda $(2x + 3y + z + 5) = 0

$ \Rightarrow $ (1 + 2$\lambda $)x + (1 + 3$\lambda $)y + (1 + $\lambda $) + (5$\lambda $ - 6) = 0

As the above plane is perpendicular to xy plane

$ \therefore $ $\overrightarrow n .\widehat k = 0$

$ \Rightarrow $ $\left[ {\left( {2 + \lambda } \right)\widehat i + \left( {3 + \lambda } \right)\widehat j + \left( {1 + \lambda } \right)\widehat k} \right].\widehat k$ = 0

$ \Rightarrow $ 1 + $\lambda $ = 0

$ \Rightarrow $ $\lambda $ = -1

So, equation of plane

- x - 2y - 11 = 0

$ \Rightarrow $ x + 2y + 11 = 0

Distance of the point (0, 0, 256) from this plane

= $\left| {{{0 + 0 + 11} \over {\sqrt 5 }}} \right|$ = ${{11} \over {\sqrt 5 }}$

Let ax + by + cz = 1 be the equation of the plane

it passed through point (0, –1, 0).

$ \therefore $ -b = 1

$ \Rightarrow $ b = -1

Also it passes through point (0, 0, 1)

$ \therefore $ c = 1

So the plane is ax - y + z = 1.

This plane an angle ${\pi \over 4}$ with the plane y – z + 5 = 0.

Normal to the plane ax - y + z = 1 is

${\overrightarrow a }$ = $a\widehat i - \widehat j + \widehat k$

Normal to the plane y – z + 5 = 0 is

${\overrightarrow b }$ = $\widehat j - \widehat k$

cos $\theta $ = $\left| {{{\overrightarrow a .\overrightarrow b } \over {\left| {\overrightarrow a } \right|\left| {\overrightarrow b } \right|}}} \right|$

$ \Rightarrow $ ${1 \over {\sqrt 2 }}$ = $\left| {{{\overrightarrow a .\overrightarrow b } \over {\left| {\overrightarrow a } \right|\left| {\overrightarrow b } \right|}}} \right|$

$ \Rightarrow $ ${{\left| {0 - 1 - 1} \right|} \over {\sqrt {{a^2} + 1 + 1} \sqrt {{1^2} + {1^2}} }}$ = ${1 \over {\sqrt 2 }}$

$ \Rightarrow $ ${a^2} + 2 = 4$

$ \Rightarrow $ a = $ \pm $ $\sqrt 2 $

$ \therefore $ Equation of plane

$ \pm $ $\sqrt 2 $x - y + z = 1

Now by checking each options you can see

equation - $\sqrt 2 $x - y + z = 1 satisfy by the point $\left( {\sqrt 2 ,1,4} \right)$.

Let ${{x - 1} \over 2} = {{y + 1} \over 3} = {{z - 2} \over 4}$ = $\lambda $

Any arbitary point on the line is P( 2$\lambda $ + 1, 3$\lambda $ - 1, 4$\lambda $ + 2).

This point also lies on the plane x + 2y + 3z = 15.

$ \therefore $ (2$\lambda $ + 1) + 2(3$\lambda $ - 1) + 3(4$\lambda $ + 2) = 15

$ \Rightarrow $ 20$\lambda $ = 10

$ \Rightarrow $ $\lambda $ = ${1 \over 2}$

So point P is (2, ${1 \over 2}$, 4).

Distance from the origin(O) of point P(2, ${1 \over 2}$, 4) is

OP = $\sqrt {{{\left( 2 \right)}^2} + {{\left( {{1 \over 2}} \right)}^2} + {{\left( 4 \right)}^2}} $

= $\sqrt {4 + {1 \over 4} + 16} $

= $\sqrt {{{81} \over 4}} $

= ${9 \over 2}$

Equation of PQ is

${{x - 2} \over {8 - 2}} = {{y + 3} \over {0 - \left( { - 3} \right)}} = {{z - 4} \over {10 - 4}}$

$ \Rightarrow $ ${{x - 2} \over 6} = {{y + 3} \over 3} = {{z - 4} \over 6}$

Point R (4, y, z) lies on this

$ \therefore $ ${{4 - 2} \over 6} = {{y + 3} \over 3} = {{z - 4} \over 6}$

$ \Rightarrow $ ${1 \over 3} = {{y + 3} \over 3} = {{z - 4} \over 6}$

y = -2 and y = 6

$ \therefore $ R = (4, -2, 6)

Distance of R(4, -2, 6) from the origin O(0, 0, 0) is

RO = $\sqrt {{4^2} + {{\left( { - 2} \right)}^2} + {6^2}} $

= $\sqrt {56} = 2\sqrt {14} $

Given,

P₁ : x + y + z = 1

P₁ : 2x + 3y + 4z = 5

Equation of the plane passing through the line of intersection of the plane P₁ and P₂ is :

P₁ + $\lambda $P₂ = 0

$ \Rightarrow $ (x + y + z –1) + $\lambda $(2x + 3y + 4z – 5) = 0

$ \Rightarrow $ x(1 + 2$\lambda $) + y(1 + 3$\lambda $) + z(1 + 4$\lambda $) - 5$\lambda $ - 1 = 0 .....(1)

Direction Ratio (D.R) of this plane = (1 + 2$\lambda $, 1 + 3$\lambda $, 1 + 4$\lambda $)

Plane (1) is perpendicular to x - y + z = 0, whose D.R = (1, -1, 1)

As they are perpendicular so dot product of D.R = 0

$ \therefore $ (1) (1 + 2$\lambda $) + (–1) (1 + 3$\lambda $) + (1) (1 + 4$\lambda $) = 0

$ \Rightarrow $ 1 + 2$\lambda $ –1 – 3$\lambda $ + 1 + 4$\lambda $ = 0

$ \Rightarrow $ $\lambda $ = $ - {1 \over 3}$

Putting the value of $\lambda $ in equation (1), we get

$ \Rightarrow $ ${x \over 3} - {z \over 3} + {2 \over 3} = 0$

$ \Rightarrow $ x - z + 2 = 0

Vector form of this plane,

$\mathop r\limits^ \to . \left( {\mathop i\limits^ \wedge - \mathop k\limits^ \wedge } \right) + 2 = 0$

Let vector $\overrightarrow p $ is perpendicular to the both vectors $\mathop {i}\limits^ \wedge + \mathop {j}\limits^ \wedge + \mathop k\limits^ \wedge $ and $\mathop {i}\limits^ \wedge + \mathop {2j}\limits^ \wedge + \mathop {3k}\limits^ \wedge $.

$ \therefore $ $\overrightarrow p $ = ($\mathop {i}\limits^ \wedge + \mathop {j}\limits^ \wedge + \mathop k\limits^ \wedge $) $ \times $ ($\mathop {i}\limits^ \wedge + \mathop {2j}\limits^ \wedge + \mathop {3k}\limits^ \wedge $)

= $\left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 1 & 1 & 1 \cr 1 & 2 & 3 \cr } } \right|$

= $\widehat i - 2\widehat j + \widehat k$

Now a vector $\overrightarrow a $ = $\mathop {2i}\limits^ \wedge + \mathop {3j}\limits^ \wedge + \mathop k\limits^ \wedge $ is given and we have to findout projection of vector $\overrightarrow a $ on $\overrightarrow p $.

$ \therefore $ Projection of vector $\overrightarrow a $ on $\overrightarrow p $

= $\left| {\overrightarrow a } \right|\cos \theta $

= $\left| {\overrightarrow a } \right| \times {{\overrightarrow a .\overrightarrow p } \over {\left| {\overrightarrow a } \right|\left| {\overrightarrow p } \right|}}$

= ${{\overrightarrow a .\overrightarrow p } \over {\left| {\overrightarrow p } \right|}}$

= ${{\left( {2\widehat i + 3\widehat j + \widehat k} \right).\left( {\widehat i - 2\widehat j + \widehat k} \right)} \over {\sqrt {1 + 4 + 1} }}$

= ${{2 - 6 + 1} \over {\sqrt 6 }}$

= ${{ - 3} \over {\sqrt 6 }}$

Magnitude of projection of vector $\overrightarrow a $ on $\overrightarrow p $

= $\left| {{{ - 3} \over {\sqrt 6 }}} \right|$ = ${{3 \over {\sqrt 6 }}}$ = ${{\sqrt 3 } \over {\sqrt 2 }}$

The equation of any plane passing through the intersection of the planes 2x – y – 4 = 0 and y + 2z – 4 = 0 is :

(2x – y – 4) + $\lambda $(y + 2z – 4) = 0 ........(1)

As this plane passes through (1, 1, 0) then this point satisfy the equation (1).

$ \therefore $ (2 – 1 – 4) + $\lambda $(1 + 0 – 4) = 0

$ \Rightarrow $ $\lambda $ = -1

Equation of required plane will be

(2x – y – 4) – (y + 2z – 4) = 0

$ \Rightarrow $ 2x – 2y – 2z = 0

$ \Rightarrow $ x – y – z = 0

Let ${{x + 3} \over {10}}$= ${{y - 2} \over {-7}}$ = ${{z} \over {1}}$ = $\lambda $

$ \therefore $ General point Q(10$\lambda $ - 3, -7$\lambda $ + 2, $\lambda $).

$\overrightarrow {PQ} $ = (10$\lambda $ - 5)$\widehat i$ + (3 - 7$\lambda $)$\widehat j$ + ($\lambda $ - 4)$\widehat k$

The vector parrallel to the given line is

= 10$\widehat i$ - 7$\widehat j$ + $\widehat k$

As $\overrightarrow {PQ} $ and (10$\widehat i$ - 7$\widehat j$ + $\widehat k$) are perpendicular to the each other.

$\overrightarrow {PQ} $ . (10$\widehat i$ - 7$\widehat j$ + $\widehat k$) = 0

$ \Rightarrow $ (10$\lambda $ - 5)10 + (3 - 7$\lambda $)(-7) + ($\lambda $ - 4)(1) = 0

$ \Rightarrow $ 100$\lambda $ – 50 + 49$\lambda $ – 21 + $\lambda $ – 4 = 0

$ \Rightarrow $ 150$\lambda $ = 75

$ \Rightarrow $ $\lambda $ = ${1 \over 2}$

$ \therefore $ The point Q = (2, $ - {3 \over 2}$, ${1 \over 2}$)

$ \therefore $ Length of perpendicular (PQ) = $\sqrt {0 + {1 \over 4} + {{49} \over 4}} $

= $\sqrt {{{50} \over 4}} = {5 \over {\sqrt 2 }}$ = 3.53

All four points are coplanar so

$\left| {\matrix{ {1 - {\lambda ^2}} & 2 & 0 \cr 2 & { - {\lambda ^2} + 1} & 0 \cr 2 & 2 & { - {\lambda ^2} - 1} \cr } } \right| = 0$

($\lambda $² + 1)² (3 $-$ $\lambda $²) = 0

$\lambda $ = $ \pm $$\sqrt 3 $

DR's of line are 2, 1, $-$2

normal vector of plane is $\widehat i$ $-$ 2$\widehat j$ $-$ k$\widehat k$

sin$\alpha $ = ${{\left( {2\widehat i + \widehat j - 2\widehat k} \right).\left( {\widehat i - 2\widehat j - k\widehat k} \right)} \over {3\sqrt {1 + 4 + {k^2}} }}$

sin $\alpha $ = ${{2k} \over {3\sqrt {{k^2} + 5} }}$         . . . . . . (1)

cos $\alpha $ =${{2\sqrt 2 } \over 3}$         . . . . .. . (2)

(1)² + (2)² = 1 $ \Rightarrow $ k² = ${5 \over 3}$

$\left| {\matrix{ i & j & k \cr 3 & 5 & 7 \cr 1 & 4 & 7 \cr } } \right|$

= $\widehat i$(35 $-$ 28) $-$ $\widehat j$(21.7) + $\widehat k$(12 $-$ 5)

= 7$\widehat i$ $-$ 14$\widehat j$ + 7$\widehat k$

= $\widehat i$ $-$ 2$\widehat j$ + $\widehat k$

1(x + 2) $-$ 2(y $-$ 2) + 1 (z + 15) = 0

x $-$ 2y + z + 11 = 0

${{11} \over {\sqrt {4 + 1 + 1} }} = {{11} \over {\sqrt 6 }}$

$\overrightarrow {OP} \times \overrightarrow {OQ} = \left( {\widehat i + 2\widehat j + \widehat k} \right) \times \left( {2\widehat i + \widehat j + 3\widehat k} \right)$

= 5$\widehat i$ $-$ $\widehat j$ $-$ 3$\widehat k$

$\overrightarrow {PQ} \times \overrightarrow {PR} = \left( {\widehat i - \widehat j + 2\widehat k} \right) \times \left( { - 2\widehat i - \widehat j + \widehat k} \right)$

= $\widehat i$ $-$ 5$\widehat j$ $-$ 3$\widehat k$

cos$\theta $ = ${{5 + 5 + 9} \over {{{\left( {\sqrt {25 + 9 + 1} } \right)}^2}}} = {{19} \over {35}}$

Point on L₁ ($\lambda $ + 3, 3$\lambda $ $-$ 1, $-$$\lambda $ + 6)

Point on L₂ (7$\mu $ $-$ 5, $-$6$\mu $ + 2, 4$\mu $ + 3

$ \Rightarrow $  $\lambda $ + 3 = 7$\mu $ $-$ 5      . . . . (i)

3$\lambda $ $-$ 1 = $-$6$\mu $ + 2       . . . .(ii)

$ \Rightarrow $  $\lambda $ = $-$1, $\mu $ = 1

point R(2, $-$ 4, 7)

Reflection is (2, $-$4, $-$ 7)

Normal vector of plane

$ = \left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 2 & { - 5} & 0 \cr 4 & { - 4} & 5 \cr } } \right|$

$ = - 4\left( {5\widehat i + 2\widehat j - 3\widehat k} \right)$

equation of plane is

5(x $-$ 7) + 2y $-$ 3(z $-$ 6) = 0

5x + 2y $-$ 3z = 17

5 $ \times $ 2 + 2$\alpha $ $-$ 3$\beta $ = 17

$ \therefore $  2$\alpha $ $-$ 3$\beta $ = 17 $-$ 10 = 7

The normal vector of required plane

$ = \left( {2\widehat i - \widehat j + 3\widehat k} \right) \times \left( {2\widehat i + 3\widehat j - \widehat k} \right)$

$ = - 8\widehat i + 8\widehat j + 8\widehat k$

So, direction ratio of normal is $\left( { - 1,1,1} \right)$

So required plane is

$ - \left( {x - 3} \right) + \left( {y + 2} \right) + \left( {z - 1} \right) = 0$

$ \Rightarrow - x + y + z + 4 = 0$

Which is satisfied by $\left( {2,0, - 2} \right)$

Let the equation of plane be

a(x $-$ 0) + b(y + 1) + c(z $-$ 0) = 0

It passes through (0, 0, 1) then

b + c = 0     . . . . (1)

Now cos ${\pi \over 4}$ = ${{a\left( 0 \right) + b\left( 1 \right) + c\left( { - 1} \right)} \over {\sqrt 2 \sqrt {{a^2} + {b^2} + {c^2}} }}$

$ \Rightarrow $  a² $=$ $-$ 2bc and b $=$ $-$ c

we get a² $=$ 2c²

$ \Rightarrow $  a $=$ $ \pm $ $\sqrt 2 $ c

$ \Rightarrow $  direction ratio (a, b, c) = ($\sqrt 2 $, $-$1, 1) or ($\sqrt 2 $, 1, $-$ 1)

General point on the given line is

x = 2$\lambda $ + 4

y = 2$\lambda $ + 5

z = $\lambda $ + 3

Solving with plane,

2$\lambda $ + 4 + 2$\lambda $ + 5 + $\lambda $ + 3 = 2

5$\lambda $ + 12 = 2

5$\lambda $ = $-$ 10

$\lambda $ = $-$ 2

p : 3(x $-$ 0) + 5(y $-$ 2) + 1 (z $-$ 5) = 0

3x + 5y + z = 15

On checking all the options, the option (4, 1, − 2) satisfy the equation of plane.

Let $\overrightarrow n $ be the normal vector to the plane passing through (4, $-$1, 2) and parallel to the lines L₁ & L₂

then $\overrightarrow n $ = $\left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 3 & { - 1} & 2 \cr 1 & 2 & 3 \cr } } \right|$

$ \therefore $  $\overrightarrow n $ = $ - 7\widehat i - 7\widehat j + 7\widehat k$

$ \therefore $  Equation of plane is

$-$ 1(x $-$ 4) $-$ 1(y + 1) + 1(z $-$ 2) = 0

$ \therefore $  x + y $-$ z $-$ 1 = 0

Now check options

Let point A is

$\left( {1 - 3\mu } \right)\widehat i + \left( {\mu - 1} \right)\widehat j + \left( {2 + 5\mu } \right)\widehat k$

and point B is (3, 2, 6)

then $\overrightarrow {AB} = \left( {2 + 3\mu } \right)\widehat i + \left( {3 - \mu } \right)\widehat j + \left( {4 - 5\mu } \right)\widehat k$

which is parallel to the the plane x $-$ 4y + 3z = 1

$ \therefore $  2 + 3$\mu $ $-$ 12 + 4$\mu $ + 12 $-$ 15$\mu $ = 0

8$\mu $ = 2

$\mu $ = ${1 \over 4}$

Vector $ \bot $ to given plane = $\left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 3 & 4 & 2 \cr 4 & 2 & 3 \cr } } \right|$

= $\widehat i\left( {12 - 4} \right) - \widehat j\left( {9 - 8} \right) + \widehat k\left( {6 - 16} \right)$

= $8\widehat i - \widehat j - 10\widehat k\,$      . . . . (1)

Vector parallel to given line

= $2\widehat i + 3\widehat j + 4\widehat k\,\,\,\,$      . . . (2)

Vector $ \bot \,\,\,$ to both (1) & (2) vector

= $\left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 8 & { - 1} & { - 10} \cr 2 & 3 & 4 \cr } } \right|$

= $\widehat i\left( { - 4 + 30} \right) - \widehat j\left( {32 + 20} \right) + \widehat k\left( {24 + 2} \right)$

= $26\widehat i - 52\widehat j + 26\widehat k$

Dr's of normal of required plane is

(26, $-$52, 26) $ \Rightarrow $ (1, $-$2, 1)

Equation of plane whose Dr's of Normal is (1, $-$2, 1) and passes through origin

1.(x $-$ 0) $-$ 2(y $-$ 0) + 1.(z $-$ 0) = 0

x $-$ 2y + z = 0

Equation of 1^st line is

${{x - b} \over a} = {y \over 1} = {{z - d} \over c}$

Dr's of 1^st line = ($a$, 1 , c)

Equation of 2^nd line is

${{x - b'} \over {a'}} = {{y - b'} \over {c'}} = {z \over 1}$

Dr's of 2^nd line = ($a'$, c' , 1)

Lines are perpendicular, so the dot product of the Dr's of two lines are zero.

$ \therefore $ $aa$' + c + c' = 0

The equation of plane

(x + y + z $-$ 1) + $\lambda $ (2x + 3y $-$ z + 4) = 0

$ \Rightarrow $  (1 + 2$\lambda $)x + (1 + 3$\lambda $)y + (1 $-$ $\lambda $)z + 4$\lambda $ $-$ 1 = 0

As plane is parallel to y axis so the normal vector of plane and dot product of $\widehat j$ is zero.

$ \therefore $  1 + 3$\lambda $ = 0

$ \Rightarrow $  $\lambda $ = $-$ ${1 \over 3}$

$ \therefore $  So the equation of the plane is

x(1 $-$ ${2 \over 3}$) + (1 $-$ ${3 \over 3}$) y + (1 + ${1 \over 3}$) $-$ ${4 \over 3}$ $-$ 1 = 0

$ \Rightarrow $  x (${1 \over 3}$) + z(${4 \over 3}$) $-$ ${7 \over 3}$ = 0

$ \Rightarrow $  x + 4z $-$ 7 = 0

By checking each options you can see only point (3, 2, 1) lies on the plane.

The line L is parallel to the plane P and intersect with line 4 at point R.

Let the coordinate of point R is, (x₁, y₁, z₁) and it passes through L₂.

${{{x_1} + 1} \over { - 3}} = {{{y_1} - 3} \over 2} = {{{z_1} - 2} \over { - 1}} = t$

$ \therefore $   x₁ = $-$1 $-$ 3t, y₁ = 3 + 2t, z₁ = 2 $-$ t

$\overrightarrow {AR} = \left( {3 - 3t} \right)\widehat i + \left( {2t} \right)\widehat j + \left( {1 - t} \right)\widehat k$

$\overrightarrow n = \widehat i + 2\widehat j - \widehat k$

As $\overrightarrow {AR} $ and $\overrightarrow n $ are perpendicular to each other, So

$\overrightarrow {AR} $ $ \cdot $ $\overrightarrow n $ = 0

$ \Rightarrow $  (3 $-$ 3t) 1 + (2t)2 + (1 $-$ t) ($-$ 1) = 0

$ \Rightarrow $  3 $-$ 3t + 4t $-$ 1 + t = 0

$ \Rightarrow $  2 + 2t = 0

$ \Rightarrow $  t = $-$ 1

$ \therefore $   point R = (2, 1, 3)

$ \therefore $  DR of line L is

= (2 $-$ ($-$ 4), $1$ $-$ 3, 3 $-$ 1)

= (6, $-$ 2, 2)

$ \therefore $  Equation of line is

${{x + 4} \over 6} = {{y - 3} \over { - 2}} = {{z - 1} \over 2}$

or ${{x + 4} \over 3} = {{y - 3} \over { - 1}} = {{z - 1} \over 1}$

Given lines,

${L_1}:r = \lambda \widehat i$, $\lambda $ $ \in $ R .......(i)

${L_2}:r = \widehat k + \mu \widehat j$, $\mu $ $ \in $ R .......(ii)

and ${L_3}:r = \widehat i + \widehat j + v\widehat k$, v $ \in $ R .......(iii)

Now, let the point P on L₁ = ($\lambda $, 0, 0)

the point Q on L₂ = (0, $\mu $, 1), and

the point R on L₃ = (1, 1, v)

For collinearity of points P, Q and R, there should be a non-zero scalar 'm', such that PQ = m PR

$ \Rightarrow ( - \lambda \widehat i + \mu \widehat j + \widehat k) = m[(1 - \lambda )\widehat i + \widehat j + v\widehat k]$

$ \Rightarrow {\lambda \over {\lambda - 1}} = {\mu \over 1} = {1 \over v}$

$ \Rightarrow v = {1 \over \mu }$ and $\lambda = {\mu \over {\mu - 1}}$

where, $\mu \ne 0$ and $\mu \ne 1$

$ \Rightarrow $ $Q \ne \widehat k$ and $Q \ne \widehat k + \widehat j$

Hence, Q can not have coordinator (0, 0, 1 ) and (0, 1, 1)

Hence, options (c) and (d) are correct.

Given lines

${L_1}:r = \widehat i + \lambda ( - \widehat i + 2\widehat j + 2\widehat k),\,\lambda \in R$ and

${L_2}:r = \mu (2\widehat i - \widehat j + 2\widehat k),\,\mu \in R$

and since line L₃ is perpendicular to both lines L₁ and L₂.

Then a vector along L₃ will be,

$\left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr { - 1} & 2 & 2 \cr 2 & { - 1} & 2 \cr } } \right| = \widehat i(4 + 2) - \widehat j( - 2 - 4) + \widehat k(1 - 4)$

$ = 6\widehat i + 6\widehat j - 3\widehat k = 3(2\widehat i + 2\widehat j - \widehat k)$ ....(i)

Now, let a general point on line L₁.

$P(1 - \lambda ,\,2\lambda ,\,2\lambda )$ and on line L₂ as $Q(2\mu - \mu ,\,2\mu )$ and let P and Q are point of intersection of lines L₁, L₃ and L₂, L₃, so direction ratio's of L₃

$(2\mu + \lambda - 1,\, - \mu - 2\lambda ,\,\,2\mu - 2\lambda )$ ....(ii)

Now, ${{2\mu + \lambda - 1} \over 2} = {{\, - \mu - 2\lambda } \over 2}\, = {{\,2\mu - 2\lambda } \over 1}$

[from Eqs. (i) and (ii)]

$ \Rightarrow \lambda = {1 \over 9}$ and $\mu = {2 \over 3}$

So, $P\left( {{8 \over 9},{2 \over 9},{2 \over 9}} \right)$ and $Q\left( {{4 \over 9}, - {2 \over 9},{4 \over 9}} \right)$

Now, we can take equation of line L₃ as $r = a + t(2\widehat i + 2\widehat j - \widehat k)$, where a is position vector of any point on line L₃ and possible vector of a are

$\left( {{8 \over 9}\widehat i + {2 \over 9}\widehat j + {2 \over 9}\widehat k} \right)$ or $\left( {{4 \over 9}\widehat i - {2 \over 9}\widehat j + {4 \over 9}\widehat k} \right)$ or $\left( {{2 \over 3}\widehat i + {1 \over 3}\widehat k} \right)$

Hence, options (a), (b) and (c) are correct.

Equation of plane passing through three given points is :

$\left| {\matrix{ {x - {x_1}} & {y - {y_1}} & {z - {z_1}} \cr {{x_2} - {x_1}} & {{y_2} - {y_1}} & {{z_2} - {z_1}} \cr {{x_3} - {x_1}} & {{y_3} - {y_1}} & {{z_3} - {z_1}} \cr } } \right| = 0$

$ \Rightarrow $  $\left| {\matrix{ {x + 2} & {y + 2} & {z - 2} \cr {1 + 2} & { - 1 + 2} & {2 - 2} \cr {1 + 2} & {1 + 2} & {1 - 2} \cr } } \right| = 0$

$ \Rightarrow $  $\left| {\matrix{ {x + 2} & {y + 2} & {z - 2} \cr 3 & 1 & 0 \cr 3 & {30} & { - 1} \cr } } \right| = 0$

$ \Rightarrow $  $ - x + 3y + 6z - 8 = 0$

$ \Rightarrow $  ${x \over 8} - {{3y} \over 8} - {{6z} \over 8} + {8 \over 8} = 0$

$ \Rightarrow $  ${x \over 8} - {y \over {{8 \over 3}}} - {z \over {{8 \over 6}}} = - 1$

$ \Rightarrow $  ${x \over { - 8}} + {y \over {{8 \over 3}}} + {z \over {{8 \over 6}}} = 1$

$ \therefore $   Sum of intercepts $ = - 8 + {8 \over 3} + {8 \over 6} = - 4$

Let $\theta $ be the angle between the two lines

Here direction cosines of ${x \over 2}$ = ${y \over 2}$ = ${z \over 1}$ are 2, 2, 1

Also second line can be written as :

${{x - 5} \over 2} = {{y - 2} \over {{P \over 7}}} = {{z - 3} \over 4}$

$ \therefore $  its direction cosines are 2, ${{P \over 7}}$, 4

Also, cos$\theta $ = ${2 \over 3}$ (Given)

$ \because $ cos$\theta $ $ = \left| {{{{a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2}} \over {\sqrt {a_1^2 + b_1^2 + c_1^2\sqrt {a_2^2 + b_2^2 + c_2^2} } }}} \right|$

$ \Rightarrow $   ${2 \over 3}$ $ = \left| {{{\left( {2 \times 2} \right) + \left( {2 \times {P \over 7}} \right) + \left( {1 \times 4} \right)} \over {\sqrt {{2^2} + {2^2} + {1^2}} \sqrt {{2^2} + {{{P^2}} \over {49}} + {4^2}} }}} \right|$

           $ = {{4 + {{2P} \over 7} + 4} \over {3 \times \sqrt {{2^2} + {{{P^2}} \over {49}} + {4^2}} }}$

$ \Rightarrow $ ${\left( {4 + {P \over 7}} \right)^2} = 20 + {{{P^2}} \over {49}}$

$ \Rightarrow $  16 + ${{8P} \over 7} + {{{P^2}} \over {49}}$ = 20 + ${{{P^2}} \over {49}}$

$ \Rightarrow $ ${{8P} \over 7} = 4$ $ \Rightarrow $ $P = {7 \over 2}$

PQ is the projection of line segment AB on the plane x + y + z = 7

P and Q are called foot of perpendicular on the plane x + y + z = 7

Let P = (x, y, z) then

${{x - 5} \over 1} = {{y + 1} \over 1} = {{z - 4} \over 1} = {{ - \left( {5 - 1 + 4} \right)} \over {{1^2} + {1^2} + {1^2}}}$

$ \Rightarrow \,\,\,\,x - 5 = y + 1 = z - 4 = - {8 \over 3}$

$\therefore\,\,\,$ x = ${7 \over 3}$ , y = $-$ ${{11} \over 3},$ z = ${4 \over 3}$

$\therefore\,\,\,$ Point P = $\left( {{7 \over 3}, - {{11} \over 3},{4 \over 3}} \right)$

Let Q = (x₁, y₁, z₁) then

${{{a_1} - 4} \over 1} = {{{y_1} + 1} \over 1} = {{{z_1} - 3} \over 1} = {{ - \left( {4 - 1 + 3} \right)} \over {{1^2} + {1^2} + {1^2}}}$

$ \Rightarrow \,\,\,\,$ x₁ $-$ 4 = y₁ +1= z₁ $-$ 3 = $-$ 2

$\therefore\,\,\,$ x = 2, y = $-$ 3, z = 1

$\therefore\,\,\,$ Point Q = (2, $-$ 3, 1)

Now length of PQ is

$\sqrt {{{\left( {{7 \over 3} - 2} \right)}^2} + {{\left( { - {{11} \over 3} + 3} \right)}^2} + {{\left( {{4 \over 3} - 1} \right)}^2}} $

$ = \sqrt {{1 \over 9} + {4 \over 9} + {1 \over 9}} $

$ = \sqrt {{6 \over 9}} $

$ = \sqrt {{2 \over 3}} $

L₁ is the line of intersection of plane 1 and plane 2.

L₂ is the line of intersection of plane $3$ and plane 4.

Line L₁ and L₂ are present on plane 5.

Now we have to find the distance of origin from the plane 5.

The plane 5 is passing through the Line L₁,

and L₁ is the line of intersection of first two planes.

$\therefore$ Equation of plane 5,

$(2x-2y+3z-2)$ + $\lambda \left( {x - y + z + 1} \right)0$

Now find a point on L₂ which is line of intersection of

x + 2y $-$ z $-$ 3 = 0 $\,\,\,\,$ .....(1)

and 3x $-$ y + 2z $-$ 1 = 0 $\,\,\,\,$ .....(2)

Put x = 0 at both (1) and (2),

2y $-$ z = 3 $\,\,\,\,.....$(3)

$-$y + 2z = 1 $\,\,\,\,....$ (4)

by solving (3) and (4) we get

$y = {7 \over 3},\,z = {5 \over 3}$

So, the point on line L₂ is $ = \left( {0,{7 \over 3},{5 \over 3}} \right)$

This point is also present on the plane $5.$

So, putting this point on equation of plane $5,$

$ - {{14} \over 5} + 5 - 2 + \lambda \left( { - {7 \over 3} + {5 \over 3}} \right) = 0$

$ \Rightarrow - {5 \over 3} + {\lambda \over 3} = 0$

$ \Rightarrow \lambda = 5$

$\therefore$ equation of plane 5 is

2x $-$ 2y + 3z $-$ 2 + $\lambda $(x $-$ y + z + 1) = 0

$ \Rightarrow $ 2x $-$ 2y + 3z $-$ 2 + 5 (x $-$ y + z + 1) = 0

$ \Rightarrow $ 7x $-$ 7y + 8z + 3 = 0

Distance of this plane 5 from (0, 0, 0)

$ = {3 \over {\sqrt {49 + 49 + 64} }}$

$ = {3 \over {\sqrt {162} }}$

$ = {3 \over {9\sqrt 2 }}$

$ = {1 \over {3\sqrt 2 }}$

Given

l + 3m + 5n = 0

and 5$l$m $-$ 2mn + 6n$l$ = 0

From eq. (1) we have

$l$ = $-$ 3m $-$ 5n

Put the value of $l$ in eq. (2), we get ;

5 ($-$3m $-$5n) m $-$ 2mn + 6n ($-$ 3m $-$ 5n) = 0

$ \Rightarrow $  15m² + 45mn + 30n² = 0

$ \Rightarrow $ m² + 3mn + 2n² = 0

$ \Rightarrow $  m² + 2mn + mn + 2n² = 0

$ \Rightarrow $ $\,\,\,$ (m + n) (m + 2n) = 0

$ \therefore $  m = $-$ n   or   m = $-$ 2n

For m = $-n, $ $l$ = $-$ 2n

And for m = $-$ 2n, $l$ = n

$ \therefore $   ($l$, m, n) = ($-$2n, $-$n, n) Or ($l$, m, n) = (n, $-$ 2n, n)

$ \Rightarrow $  ($l$, m, n) = ($-$2, $-$1, 1) Or ($l$, m, n) = (1, $-$ 2, 1)

Therefore, angle between the lines is given as :

cos ($\theta $) = ${{\left( { - 2} \right)\left( 1 \right) + \left( { - 1} \right).\left( { - 2} \right) + \left( 1 \right)\left( 1 \right)} \over {\sqrt 6 .\sqrt 6 }}$

$ \Rightarrow $ cos ($\theta $) = ${1 \over 6}$ $ \Rightarrow $  $\theta $ =cos^$-$1 $\left( {{1 \over 6}} \right)$

Since the plane bisects the line joining the points (1, 2, 3) and ($-$3, 4, 5) then the plane passes through the midpoint of the line which is :

$\left( {{{1 - 3} \over 2},{{2 + 4} \over 2},{{5 + 3} \over 2}} \right)$  $ \equiv $  $\left( {{{ - 2} \over 2},{6 \over 2},{8 \over 2}} \right)$ $ \equiv $ ($-$1, 3, 4).

As plane cuts the line segment at right angle, so the direction cosines of the normal of the plane are ($-$ 3 $-$ 1, 4 $-$ 2, 5 $-$ 3) = ($-$ 4, 2, 2)

So the equation of the plane is : $-$ 4x + 2y + 2z = $\lambda $

As plane passes through ($-$ 1, 3, 4)

So, $-$ 4($-$ 1) + 2(3) + 2(4) = $\lambda $ $ \Rightarrow $ $\lambda $ = 18

Therefore, equation of plane is : $-$ 4x + 2y + 2z = 18

Now, only ($-$ 3, 2, 1) satiesfies the given plane as

$-$ 4($-$ 3) + 2(2) + 2(1) = 18

If a, b, c are the intercepts of the variable plane on the x,y,z axes respectively, then the equation of the plane is

${x \over a} + {y \over b} + {z \over c} = 1$

And the point of intersection of the planes parallel to the xy, yz and zx planes is $\left( {a,b,c} \right)$.

As the point (3, 2, 1) lies on the variables plane,

so ${3 \over a} + {2 \over b} + {1 \over c} = 1$

Therefore, the required locus is

${3 \over x} + {2 \over y} + {1 \over z} = 1$

Normal to $3x + 4y + z = 1$    is $3\widehat i + 4\widehat j + \widehat k$

Normal to $5x + 8y + 2z =$ $ - 14$ is $5\widehat i + 8\widehat j + 2\widehat k$

The line of intersection of the planes is perpendicular to both normals, so, direction ratios of the intersection line are directly proportional to the cross product of the normal vectors.

Therefore the direction ratios of the line is $ - \widehat j + 4\widehat k$

Hence the angle between the plane x + y + z + 5 = 0

and the intersection line is ${\sin ^{ - 1}}\left( {{{ - 1 + 4} \over {\sqrt {17} \sqrt 3 }}} \right) = {\sin ^{ - 1}}\left( {\sqrt {{3 \over {17}}} } \right)$

We have,

and $\eqalign{ & {P_1}:2x + y - z = 3 \cr & {P_2}:x + 2y + z = 2 \cr} $

Here, $\overrightarrow {{n_1}} = 2\widehat i + \widehat j - \widehat k$

and $\overrightarrow {{n_2}} = \widehat i + 2\widehat j + \widehat k$

(a) Direction ratio of the line of intersection of P₁ and P₂ is $\theta $ $\overrightarrow {{n_1}} \times \overrightarrow {{n_2}} $

i.e. $\left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 2 & 1 & { - 1} \cr 1 & 2 & 1 \cr } } \right|$

$ = (1 + 2)\widehat i - (2 + 1)\widehat j + (4 - 1)\widehat k$

$ = 3(\widehat i - \widehat j + \widehat k)$

Hence, statement a is false.

(b) We have,

${{3x - 4} \over 9} = {{1 - 3y} \over 9} = {z \over 3}$

$ \Rightarrow {{x - {4 \over 3}} \over 3} = {{\left( {y - {1 \over 3}} \right)} \over { - 3}} = {z \over 3}$

This line is parallel to the line of intersection of P₁ and P₂.

Hence, statement (b) is false.

(c) Let acute angle between P₁ and P₂ be $\theta $.

We know that,

$\cos \theta = {{\overrightarrow {{n_1}} .\overrightarrow {{n_2}} } \over {|\overrightarrow {{n_1}} ||\overrightarrow {{n_2}} |}}$

$ = {{(2\widehat i + \widehat j - \widehat k).(\widehat i + 2\widehat j + \widehat k)} \over {|2\widehat i + \widehat j - \widehat k||\widehat i + 2\widehat j + \widehat k|}}$

$ = {{2 + 2 - 1} \over {\sqrt 6 \times \sqrt 6 }} = {1 \over 2}$

$\theta = 60^\circ $

Hence, statement (c) is true.

(d) Equation of plane passing through the point (4, 2, $-$2) and perpendicular to the line of intersection of P₁ and P₂ is

$\eqalign{ & 3(x - 4) - 3(y - 2) + 3(z + 2) = 0 \cr & \Rightarrow 3x - 3y + 3z - 12 + 6 + 6 = 0 \cr} $

$ \Rightarrow 3x - 3y + 3z - 12 + 6 + 6 = 0$

$ \Rightarrow x - y + z = 0$

Now, distance of the point (2, 1, 1) from the plane x $-$ y + z = 0 is

$D = \left| {{{2 - 1 + 1} \over {\sqrt {1 + 1 + 1} }}} \right| = {2 \over {\sqrt 3 }}$

Hence, statement (d) is true.

Suppose centroid be (h, k, $l$)

$ \therefore $   x $-$ intp $=$ 3h, y $-$ intp $=$ 3k, z $-$ intp $=$ 3$l$

Equation ${x \over {3h}} + {y \over {3k}} + {z \over {3l}} = 1$

$ \therefore $   Distance from (0, 0, 0)

$\left| {{{ - 1} \over {\sqrt {{1 \over {9{h^2}}} + {1 \over {9{k^2}}} + {1 \over {9{l^2}}}} }}} \right| = 3$

$ \Rightarrow $   ${1 \over {{x^2}}} + {1 \over {{y^2}}} + {1 \over {{z^2}}} = 1$

Point (3, $-$ 2, $-$ $\lambda $) on p line 2x $-$ 4y + 3z $-$ 2 $=$ 0

$=$ 6 + 8 $-$ 3$\lambda $ $-$ 2 = 0

$=$ 3$\lambda $ $=$ 12

$\lambda $ $=$ 4

Now,

${{x - 3} \over 1} = {{y + 2} \over { - 1}} = {{z + 4} \over { - 2}} = {k_1}$          . . .(i)

${{x - 1} \over {12}} = {y \over 9} = {z \over 4} = {k_2}$                 . . .(ii)

Point on equation (i) P (k₁ + 3, $-$ k₁ $-$ 2, $-$ 2k₁ $-$ 4)

Point on equation (ii) Q(12k₂ + 1, 9k₂, 4k₂)

k₁ + 3 $=$ 12k₂ + 1 $\left| { - {k_1} - 2 = 9{k_2}} \right|$ $-$ 2k₁ $-$ 4 $=$ 4k₂

k₂  $=$  0

k₁   $=$  $-$ 2

p (1, 0, 0) lie on equation of a line 1

gives shortest distance $=$ 0

Given,

x + 8y + 7z = 0

9x + 2y + 3z = 0

x + y + z = 0

Solving those equations, we get

x = $\lambda $, y = 6$\lambda $, z = -7$\lambda $

This point lies on the plane x + 2y + z = 6

$ \therefore $ $\lambda $ + 2(6$\lambda $) + (-7$\lambda $) = 0

$ \Rightarrow $ $\lambda $ = 1

$ \therefore $ x = 1, y = 6, z = -7

$ \therefore $ a = 1, b = 6, c = -7

So, 2a + b + c = 2(1) + 6 + (-7) = 1

$\overrightarrow n = \overrightarrow {{n_1}} \times \overrightarrow {{n_2}} $

$ \Rightarrow $   $\left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 3 & { - 1} & 1 \cr 1 & 4 & { - 2} \cr } } \right| = \widehat i\left( { - 2} \right) - \widehat j\left( { - 7} \right) + \widehat k\left( {13} \right)$

$\overrightarrow n = - 2\widehat i + 7\widehat j + 13\widehat k$

Now,

        $3x - y + z = 1$

        $x + 4y - 2z = 2$

but   z $=$ 0 & solving the given

        x $=$ 6/13 & y = 5/13

$ \therefore $   required equation of a line is

${{x - 6/13} \over 2} = {{y - 5/13} \over { - 7}} = {z \over { - 13}}$

$\overrightarrow n $ = $\overrightarrow {{n_1}} \times \overrightarrow {{n_2}} $ = $\left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 6 & 7 & 8 \cr 3 & 5 & 7 \cr } } \right|$

= (9, $-$ 18, 9)

= (1, $-$2, 1)

$ \therefore $   Equation of plane is

1(x + 1) $-$ 2(y $-$ 1) + (z $-$ 3) = 0

$ \Rightarrow $   x $-$ 2y + z = 0

foot to z

${{x - 1} \over 1}$ = ${{y + 2} \over { - 2}}$ = ${{z - 1} \over 1}$ = $ - {{\left[ {1 + 4 + 1} \right]} \over 6}$

x = 0,   y = 0,   z = 0

Let the plane be

a(x $-$ 1) + b(y + 1) + c (z + 1) = 0

Normal vector

$\left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 1 & { - 2} & 3 \cr 2 & { - 1} & { - 1} \cr } } \right| = 5\widehat i + 7\widehat j + 3\widehat k$

So plane is 5(x $-$ 1) + 7(y + 1) + 3(z + 1) = 0

$ \Rightarrow $   5x + 7y + 3z + 5 = 0

Distance of point (1, 3, $-$ 7) from the plane is

${{5 + 21 - 21 + 5} \over {\sqrt {25 + 49 + 9} }} = {{10} \over {\sqrt {83} }}$

Equation of line PQ is ${{x - 1} \over 1} = {{y + 2} \over 4} = {{z - 3} \over 5}$

Let F be ($\lambda $ + 1, 4$\lambda $ $-$ 2, 5$\lambda $ + 3)


Since F lies on the plane

$ \therefore $   2($\lambda $ + 1) + 3(4$\lambda $ $-$ 2) $-$ 4(5$\lambda $ + 3) + 22 $=$ 0

$ \Rightarrow $   $-$ 6$\lambda $ + 6 = 0 $ \Rightarrow $  $\lambda $ = 1

$ \therefore $   F is (2, 2, 8)

PQ = 2 PF = 2$\sqrt {{1^2} + {4^2} + {5^2}} $ = 2$\sqrt {42} $

Let the equation of plane be ax + by + cz = 1. Then

a + b + c = 1

2a + b $-$ 2c = 0

3a $-$ 6b $-$ 2c = 0

$ \Rightarrow $ a = 7b

c = ${{15b} \over 2}$

b = ${{2} \over 31}$, a = ${{14} \over 31}$, c = ${{15} \over 31}$

$ \therefore $ 14x + 2y + 15z = 31

As lines are coplanar, so

$\left| {\matrix{ {3 - 1} & {2 - 2} & {1 - \left( { - 3} \right)} \cr 1 & 2 & {{\lambda ^2}} \cr 1 & {{\lambda ^2}} & 2 \cr } } \right| $ = 0

$ \Rightarrow $   $\left| {\matrix{ 2 & 0 & 4 \cr 1 & 2 & {{\lambda ^2}} \cr 1 & {{\lambda ^2}} & 2 \cr } } \right| $ = 0

$ \Rightarrow $   2(4 $-$ $\lambda $⁴) + 4($\lambda $² $-$ 2) = 0

$ \Rightarrow $   4 $-$ $\lambda $⁴ + 2$\lambda $² $-$ 4 = 0

$ \Rightarrow $   $\lambda $²($\lambda $² $-$ 2) = 0

$ \Rightarrow $   $\lambda $ = 0, $\sqrt 2 , - \sqrt 2 $

$ \therefore $   3 distinct real values are possible.

DR's of AD are

${{\lambda - 1} \over 2} - 2,{{5 + 3} \over 2} - 3,{{\mu + 2} \over 2} - 5$

i.e.  ${{\lambda - 5} \over 2},\,\,1,\,\,{{\mu - 8} \over 2}$

As medium is making equal angles with coordinate axes,

$ \therefore $   ${{\lambda - 5} \over 2} = 1 = {{\mu - 8} \over 2}$

$ \Rightarrow $   $\lambda $ = 7,   $\mu $ = 10

$ \therefore $   $\lambda $³ + $\mu $³ + 5 = 7³ + 10³ + 5 = 1348

Shortest distance between the lines

${{x - {x_1}} \over {{a_1}}} = {{y - {y_1}} \over {{b_1}}} = {{z - {z_1}} \over {{c_1}}}$

and  ${{x - {x_2}} \over {{a_2}}} = {{y - {y_2}} \over {{b_2}}} = {{z - {z_2}} \over {{c_2}}}$   is

   $\left| {{{\left| {\matrix{ {{x_2} - {x_1}} & {{y_2} - {y_1}} & {{z_2} - {z_1}} \cr {{a_1}} & {{b_1}} & {{c_1}} \cr {{a_2}} & {{b_2}} & {{c_2}} \cr } } \right|} \over {\sqrt {{{\left( {{b_1}{c_2} - {b_2}{c_1}} \right)}^2} + {{\left( {{c_1}{a_2} - {c_2}{a_1}} \right)}^2} + {{\left( {{a_1}{b_2} - {a_2}{b_1}} \right)}^2}} }}} \right|$

$ \therefore $   Shortest distance between two given lines are,

   $\left| {{{\left| {\matrix{ { - 2} & 4 & 5 \cr 2 & 2 & 1 \cr { - 1} & 8 & 4 \cr } } \right|} \over {\sqrt {{{\left( {8 - 8} \right)}^2} + {{\left( { - 1 - 8} \right)}^2} + {{\left( {16 + 2} \right)}^2}} }}} \right|$

=   $\left| {{{ - 36 + 90} \over {\sqrt {405} }}} \right|$

=   ${{54} \over {20.1}}$

=   2.68