Given w(x, y, z) = \(\left( \frac { { 5x }^{ 3 }{ y }^{ 4 }+7{ y }^{ 2 }{ xz }^{ 4 }-{ 75y }^{ 3 }{ z }^{ 4 } }{ { x }^{ 2 }+{ y }^{ 2 } } \right) \)
Let (x, y, z) = \(\frac { { 5x }^{ 3 }{ y }^{ 4 }+7{ y }^{ 2 }{ xz }^{ 4 }-{ 75y }^{ 3 }{ z }^{ 4 } }{ { x }^{ 2 }+{ y }^{ 2 } } \)
⇒ w = log f  ...(1)
⇒ e^w = f
f(λx, λy, λz) = \(\frac { { 5\lambda }^{ 3 }{ x }^{ 3 }{ \lambda }^{ 4 }{ y }^{ 4 }+7{ \lambda }^{ 2 }{ y }^{ 2 }\lambda x{ \lambda }^{ 4 }{ z }^{ 4 }-75{ \lambda }^{ 3 }{ y }^{ 3 }{ \lambda }^{ 4 }{ z }^{ 4 }{ }^{ } }{ { \lambda }^{ 2 }{ x }^{ 2 }+{ \lambda }^{ 2 }{ y }^{ 2 } } \)
= \(\frac { { \lambda }^{ 7 }(5{ x }^{ 3 }{ y }^{ 4 }+7{ y }^{ 2 }{ xz }^{ 4 }-{ 75 }y^{ 3 }{ z }^{ 4 } }{ { \lambda }^{ 2 }({ x }^{ 2 }+{ y }^{ 2 }) } ={ \lambda }^{ 5 }f(x,y,z)\)
∴ f(x, y, z) is a homogeneous function of degree 5.
∴ By Euler's theorem,
\(x.\frac { \partial f }{ \partial x } +y\frac { \partial f }{ \partial y } +z\frac { \partial f }{ \partial z } =5.f\)
⇒ \(x.\frac { \partial }{ \partial x } ({ e }^{ w })+y.\frac { \partial }{ \partial y } ({ e }^{ w })+z.\frac { \partial }{ \partial z } ({ e }^{ w })=5.{ e }^{ w }\) [using (1)]
⇒ \(x.{ e }^{ w }\frac { \partial w }{ \partial x } +y.{ e }^{ w }\frac { \partial w }{ \partial y } +z.{ e }^{ w }\frac { \partial w }{ \partial z } ({ e }^{ w })=5{ e }^{ w }\)
⇒ \(x\frac { \partial w }{ \partial x } +y\frac { \partial w }{ \partial y } +z\frac { \partial w }{ \partial z } \)  [Divided by e^w]

W(x, y, z) = xy + yz + zx, x =u -v, y = uv, z = u + v; y = uv; z = u
\(\frac { \partial W }{ \partial x } \) = y + z; \(\frac { \partial W }{ \partial y } \) = x + z
∴ \(\frac { \partial W }{ \partial x } \) = uv + u +  v;
\(\frac { \partial W }{ \partial y} \) = u - v + u + v;
\(\frac { \partial W }{ \partial z} \) = uv +  u - v
\(\frac { dx }{ du } =1;\frac { dy }{ du } =v;\frac { dz }{ du } =1\)
\(\frac { dx }{ dv } =1;\frac { dy }{ dv } =v;\frac { dz }{ dv} =1\)
By chain rule
\(\frac { \partial W }{ \partial u } =\frac { \partial w }{ \partial x } .\frac { dx }{ du } +\frac { \partial w }{ \partial y } .\frac { dy }{ du } +\frac { \partial w }{ \partial z } .\frac { dz }{ du } \)
= (uv +u +v) (1) +2u (v) + (uv +u - v)(1)

\(\frac { \partial W }{ \partial u } \) = 4uv + 2u = 12u (2v + 1)
\({ \left( \frac { \partial W }{ \partial u } \right) }_{ \left( \frac { 1 }{ 2 } ,1 \right) }\) = 2 x \(\frac12\) (2+ 1) = 1(2+ 1) = 3
= (uv + u + v) (-1) + (2u) (u) + (uv + u - v)(1)

= 2u² - 2v = 2 (u² - v)
∴  \({ \left( \frac { \partial W }{ \partial v } \right) }_{ \left( \frac { 1 }{ 2 } ,1 \right) }\) = \(2\left( \frac { 1 }{ 4 } -1 \right) =2\left( -\frac { 3 }{ 4 } \right) =-\frac { 3 }{ 2 } \)

Given z(x, y) = x³ - 3x²y³ ; x = se^t; y = se^-t
\(\frac { \partial z }{ \partial s } \) = 3x² - 6xy³; \(\frac { \partial z }{ \partial y } \) = -9x²y²
∴ \(\frac { \partial z }{ \partial s } \) 3s²e^2t - 6 se^t s³ e^-3t
\(\frac { \partial z }{ \partial y } \) = 9s²e^2t.s²e^-2t
\(\frac { \partial z }{ \partial y } \) = -9s⁴
\(\frac { \partial z }{ \partial s} \) = e^t; \(\frac { \partial y }{ \partial s} \) = e^-t
\(\frac { \partial x }{ \partial t} \) = se^t; \(\frac { \partial y }{ \partial s} \) = -se^-t 
By chain rule;
\(\therefore \frac { \partial z }{ \partial s } =\frac { \partial z }{ \partial x } .\frac { dx }{ ds } +\frac { \partial u }{ \partial y } .\frac { dy }{ dx } \)
= (3s²e^2t - 6 s⁴ e^-2t) (e^t) + (-9s⁴) (e^-t)
= 3s²e^3t - 6 s⁴ e^-t -9s⁴ e^-t
\(\frac { \partial z }{ \partial s } \) = 3s²e^3t - 15 s⁴ e^-t
By chain rule;
\(\frac { \partial z }{ \partial t } =\frac { \partial z }{ \partial x } .\frac { dx }{ dt } +\frac { \partial z }{ \partial y } .\frac { dy }{ dt } \)
= (3s²e^2t - 6 s⁴e-^2t) (se^t) + (-9 s⁴) (-se^-t)
= 3s³e^3t - 6 s⁵e^-t + 9 s⁵ e^-t
= 3s³ e^3t + 3 s⁵e^-t
= 3s³(e^3t + s²e^-t)

Given U (x, y) = e^x sin y ; x = st² ; y = s²t
\(\frac { \partial U }{ \partial x } \) = e^x sin y ; \(\frac { \partial U }{ \partial y } \) = e^x cos y
\(\frac { \partial U }{ \partial x } \) = \({ e }^{ { st }^{ 2 } }\) sin (s²t)
\(\frac { \partial U }{ \partial y } \) = \({ e }^{ { st }^{ 2 } }\) cos (s²t)
\(\frac{dx}{dt}\) = 2st; \(\frac{dy}{dt}\) = s²
\(\frac{dx}{ds}\) = t²; \(\frac{dy}{ds}\) = 2 st
By chain rule
\(\frac { dU }{ ds } =\frac { \partial U }{ \partial x } .\frac { dx }{ ds } +\frac { \partial U }{ \partial y } .\frac { dy }{ ds } \)
= \({ e }^{ { st }^{ 2 } }\). sin (s²t) (t²) + \({ e }^{ { st }^{ 2 } }\) cos(s^2t).(2^st)
∴ \({ \left( \frac { \partial U }{ \partial s } \right) }_{ (s=t=1) }\) = e¹ sin (1) + 2e¹ cos (1)
= e [sin (1) + 2 cos (1)] and
\(\frac { dU }{ dt } =\frac { \partial u }{ \partial x } .\frac { dx }{ dt } +\frac { \partial u }{ \partial y } .\frac { dy }{ dt } \)
=  \({ e }^{ { st }^{ 2 } }\) . sin (s²t)(2^st) + \({ e }^{ { st }^{ 2 } }\) cos (s²t). (s²)
∴ \({ \left( \frac { \partial U }{ \partial t } \right) }_{ (s=t=1) }\) = 2e¹ sin (1) + e¹ cos (1)
= e [2 sin (1) + cos (1)]

Let X be the random variablc denotes number of red balls.
Then X take the values 0, 1, 2
Sample space = ⁷C₂ = 21
Let X denote the drawing the red ball.
Then X take the values 0, 1, 2
P(X = 0), X^-1 (BB) = ³C₂ = 3
P(X = 1), X^-1 (BR) = ³C₁ x ⁴C₁ = 12
P(X = 2), X^-1 (BR) = ³C₂ = 6

Values of random variable	0	1	2	Total
Number of elements in inverseimage	3	12	6	21

The probability mass function is

x	0	1	2
f(x)	\(\cfrac { 1 }{ 7 } \)	\(\cfrac { 4 }{ 7 } \)	\(\cfrac { 2 }{ 7 } \)

Mean :
\(E(x)=\Sigma x.f\left( x \right) \) 
= \(0(\frac { 1 }{ 7 } )+1\left( \frac { 4 }{ 7 } \right) +2\left( \frac { 2 }{ 7 } \right) \) 
= \(\frac { 4 }{ 7 } +\frac { 4 }{ 7 } =\frac { 8 }{ 7 }\)

Given
\(f(x)=\begin{cases} \begin{matrix} { Ke }^{ \frac { -x }{ 3 } } & \begin{matrix} for & x>0 \end{matrix} \end{matrix} \\ \begin{matrix} 0 & \begin{matrix} for & x\le 0 \end{matrix} \end{matrix} \end{cases}\)
(i) Since f(x) is a probability density function
\(\int _{ -\infty }^{ \infty }{ f(x) } dx=1\) 
\(\Rightarrow \int _{ 0 }^{ \infty }{ K.{ e }^{ \frac { -x }{ 3 } } } dx=1\Rightarrow k\frac { \left[ { e }^{ \frac { -x }{ 3 } } \right] ^{ \infty } }{ -\frac { 1 }{ 3 } } \) 
\(\Rightarrow -3k\left[ { e }^{ -\infty }-{ e }^{ 0 } \right] =1\) [∵ e^∞ = 0, e⁰ = 1]
\(\Rightarrow 3k=1\Rightarrow k=\frac { 1 }{ 3 } \) 
\(\therefore k=\cfrac { 1 }{ 3 } \) 
(ii) The distribution function F(x) = \(\int _{ -\infty }^{ x }{ f(u)du } \)
Case 1: x < 0, 
\(F(x)=\int _{ -\infty }^{ x }{ f(x)dx=0 } \) 
Case 2: x > 0, 
\(f(x)=\int _{ -\infty }^{ x }{ f(x)dx } \) 
= \(\int _{ -\infty }^{ 0 }{ f(x)dx+\int _{ 0 }^{ x }{ f(x) } dx } \) 
= \(0+k\int _{ 0 }^{ x }{ { e }^{ \frac { -x }{ 3 } } } dx\) 
= \(\frac { 1 }{ 3 } \left[ \cfrac { { e }^{ \frac { -x }{ 3 } } }{ -\frac { 1 }{ 3 } } \right] =-\left[ { e }^{ -\frac { x }{ 3 } }-{ e }^{ o } \right] \) 
= \(-[{ e }^{ -\frac { x }{ 3 } }-1]\) 
= \(1-{ e }^{ -\frac { x }{ 3 } }\) 
\(\therefore F(x)=\begin{cases} \begin{matrix} 0 & x\le 0 \end{matrix} \\ \begin{matrix} 1-{ e }^{ \frac { -x }{ 3 } } & x>0 \end{matrix} \end{cases}\) 
(iii) p(X < 3)
= \(\int _{ 0 }^{ 3 }{ ke^{ -\frac { x }{ 3 } } } dx=\cfrac { 1 }{ 3 } \int _{ 0 }^{ 3 }{ { e }^{ -\frac { x }{ 3 } }dx } \)
= \(\cfrac { 1 }{ 3 } \left[ \frac { { e }^{ -\frac { x }{ 3 } } }{ \frac { -1 }{ 3 } } \right] \) 
= -[e^-1-e⁰] = -[e^-1-1]
(iv) \(p(5\le X)=p(X\ge 5)=\int _{ 5 }^{ \infty }{ f(x)dx } \) 
= \(\int _{ 5 }^{ \infty }{ ke^{ -\frac { x }{ 3 } } } dx=\cfrac { 1 }{ 3 } \cfrac { \left[ { e }^{ -\frac { x }{ 3 } } \right] _{ 5 }^{ \infty } }{ \frac { -1 }{ 3 } } \) 
= \(-\left[ { e }^{ -\infty }-e^{ \frac { -3 }{ 5 } } \right] =\left[ 0-{ e }^{ \frac { -5 }{ 3 } } \right] \) 
= \({ e }^{ \frac { -5 }{ 3 } }\) 
(v) \(p(X\le 4)=\int _{ -\infty }^{ 4 }{ f(x)dx } \) 
= \(\int _{ -\infty }^{ 0 }{ f(x)dx+\int _{ 0 }^{ 4 }{ f(x)dx } } \) 
= \(0+\int _{ 0 }^{ 4 }{ { ke }^{ -\frac { x }{ 3 } }dx } =k\left[ \frac { { e }^{ -\frac { x }{ 3 } } }{ -\frac { 1 }{ 3 } } \right] _{ 0 }^{ 4 }\) 
= \(\cfrac { 1 }{ 3 } \cfrac { \left[ { e }^{ -\frac { x }{ 3 } } \right] _{ 0 }^{ 4 } }{ -\frac { 1 }{ 3 } } =-\left[ { e }^{ \frac { -4 }{ 3 } }-{ e }^{ o } \right] \) 
= \(-\left[ { e }^{ \frac { -4 }{ 3 } }-1 \right] =1-{ e }^{ \frac { -4 }{ 3 } }\)

Given \(\begin{cases} \begin{matrix} k & 200\le x\le 600 \end{matrix} \\ \begin{matrix} 0 & otherwise \end{matrix} \end{cases}\) 
(i) Since f{x) is a probability density function
\(\int _{ -\infty }^{ \infty }{ f(x)d=1\Rightarrow \int _{ 200 }^{ 600 }{ kda } =1 } \) 
\(\Rightarrow k[x]_{ 200 }^{ 600 }=1\Rightarrow k(600-200)=1\) 
400 k = 1
\(\Rightarrow k=\frac { 1 }{ 400 } \) 
(ii) The distribution function  
= \(\int _{ -\infty }^{ x }{ f(u) } du\) 
Case 1: x < 200
\(F(x) =\int _{ -\infty }^{ u }{ du } =0\) 
Case 1: x < 200 ≤ x ≤  600
\(\int _{ -\infty }^{ x }{ f(u) } du\)
\(F(x)=\int _{ -\infty }^{ 200 }{ f(u)du } =+\int _{ 200 }^{ x }{ f(u)du } \)  
= \( =\frac { 1 }{ 400 }(x-200) =\frac { x }{ 400 } =\frac { 1 }{ 2 } \) 
Case 3: x > 600
\(F(x) =\int _{ -\infty }^{ u }{ du } =0\) 
\(f(x)= \begin{cases}0, & x<200 \\ \frac{x}{400}-\frac{1}{2}, & 200 \leq x \leq 600 \\ 0, & x>600\end{cases}\)
(iii) P(300 < x < 500)
= \(\int _{ 300 }^{ 500 }{ kdx=\frac { 1 }{ 400 } \left[ x \right] _{ 300 }^{ 500 } } \) 
= \(\frac { 1 }{ 400 } \left[ 500-300 \right] =\frac { 200 }{ 400 } =\frac { 1 }{ 2 } \)

Given

The random variable X take the values -1, 0, 1, 2, 3
For a discrete random variable X, we have
f(x) = p(X = x)
∴ f(-1) = p(X= -1) = F(-1) -F(0)
= 0.15-0 = 0.15
f(0) = p(X = 0) = F(0)-F(-1)
= 0.35-0.15 = 0.20
f(1) = p(X = 1) = F(1)-F(0)
= 0.60-0.35 = 0.25
f(2) = p(X=2) = F(2)-F(1)
= 0.85-0.60 = 0.25
f(3) = p(X = 3) = F = (3)-F(2)
= 1-0.85 = 0.15
(i) ஃThe probability mass function is

x	-1	0	1	2	3
f(x)	0.15	0.20	0.25	0.25	0.15

(ii) p(X<1)
= p(X = -1) + p(X = 0)
= 0.15 + 0.20 = 0.35
(iii) p(X ≥ 2)
= p(X = 2) + p(X = 3)
= 0.25 + 0.15
= 0.40

Given
\(\\ \\ \\ \\ \\ f(x)=\begin{cases} \begin{matrix} \frac { { x }^{ 2 }+1 }{ k } & forx=0,1,2 \end{matrix} \\ \begin{matrix} 0 & otherwise \end{matrix} \end{cases}\\ \\ \\ \\ \\ \\ \)
The random variable X take the values 0, 1, 2.
Probability mass function.

x	0	1	2
f(x)	\(\cfrac { 1 }{ k } \)	\(\cfrac { 2 }{ k } \)	\(\cfrac {5 }{ k } \)

\(\sum _{ i=0 }^{ 2 }{ f(x_{ i })=1\Rightarrow f(0)+f(1)+f(2)=1 } \)
\(\Rightarrow \frac { 0+1 }{ k } +\frac { 1+1 }{ k } +\frac { 4+1 }{ k } \) 
\(\Rightarrow \frac { 1 }{ k } +\frac { 2 }{ k } +\frac { 5 }{ k } =1\) 
\(\Rightarrow \frac { 8 }{ k } =1\) 
\(\Rightarrow k=8\) 
(ii) Cumulative distribution function

x	0	1	2
f(x)	\(\cfrac { 1 }{ 8 } \)	\(\cfrac { 2 }{ 8 } \)	1

\(F(0)=P(X<0)\\P(x=0)\\ =\frac { 1 }{ 8 } \) 
\(F(1)=P(X = 0)+ P(X = 1)\\ \frac { 1 }{ 8 } +\frac { 2 }{ 8 } =\frac { 3 }{ 8 } \)
\(F(2)=P(X= 0) + P(X = 1) + P(X = 2) = \frac { 1 }{ 8 } +\frac { 2 }{ 8 } +\frac { 5 }{ 8 } =1\) 
Cumulative distribution function is
\(f(x)=\begin{cases} \begin{matrix} \frac { 1 }{ 8 } & for & x\le 0 \end{matrix} \\ \begin{matrix} \frac { 2 }{ 8 } +\frac { 1 }{ 8 } & for & \frac { 3 }{ 8 } forx\le 1 \end{matrix} \\ \begin{matrix} \frac { 3 }{ 8 } +\frac { 5 }{ 8 } =1 & for & x\le 2 \end{matrix} \end{cases}\)
(iii) p(x ≥ 1) = p(x = 1) + p(x = 2)
= \(\frac { 2 }{ 8 } +\frac { 5 }{ 8 } \)
\(p(x\ge 1)=\frac { 7 }{ 8 } \)

Let X be the thrown random variable denotes the total in two the thrown a die.
Sample space S

I/II	1	3	3	5	5	5
1	2	4	4	6	6	6
3	4	6	6	8	8	8
3	4	6	6	8	8	8
5	6	8	8	10	10	10
5	6	8	8	10	10	10
5	6	8	8	10	10	10

n (S) = 36
X = {2, 4, 6, 8, 10}

Values of the random variable	2	4	6	8	10	Total
No. of elements in inverse images	1	4	10	12	9	36

\(p(x=2)=\cfrac { 1 }{ 36 } \)
\(p(x=4)=\cfrac { 4 }{ 36 } \)
\(p(x=6)=\cfrac { 10 }{ 36 } \)
\(p(x=8)=\cfrac { 12 }{ 36 } \)
\(p(x=10)=\cfrac { 9 }{ 36 } \)
(i) Probability mass function is

x	2	4	6	8	10
f(x)	\(\\ \cfrac { 1 }{ 36 } \)	\(\cfrac { 4 }{ 36 } \)	\(\cfrac { 10 }{ 36 } \)	\(\cfrac { 12 }{ 36 } \)	\(\cfrac { 9 }{ 36 } \)

(ii) Cumulative distribution function .
F(x) = p(X ≤ x) = \(\sum_{x_i ≤ x }\)(X = x_i)
P(X<2) = 0 for \(\infty\) < x < 2
\(F(2)=\frac { 1 }{ 36 } \)
\(F(4)=\frac { 1 }{ 36 } +\frac { 4 }{ 36 } =\frac { 5 }{ 36 } \)
\(F(6)=\frac { 1 }{ 36 } +\frac { 4 }{ 36 } +\frac { 10 }{ 36 } =\frac { 15 }{ 36 } \)
\(F(8)=\frac { 1 }{ 36 } +\frac { 4 }{ 36 } +\frac { 10 }{ 36 } +\frac { 12 }{ 36 } =\frac { 27 }{ 36 }\)
\(F(10)=\frac { 27 }{ 36 } +\frac { 9 }{ 36 } =\frac { 36 }{ 36 } =1\)
∵ The cumulative distribution function n
\(F(x)=\left\{\begin{array}{lll} 0 & \text { for } & x<2 \\ \frac{1}{36} & \text { for } & x \leq 2 \\ \frac{5}{36} & \text { for } & x \leq 6 \\ \frac{15}{36} & \text { for } & x \leq 8 \\ 1 & \text { for } & x \leq 10 \end{array}\right.\)
(iii) p(4≤ X < 10) = p(x = 4) + p(x = 6) + p(x = 8)
= \(\frac { 4 }{ 36 } +\frac { 10 }{ 36 } +\frac { 12 }{ 36 } +\frac { 26 }{ 36 } =\frac { 13 }{ 18 } \)  
(iv) p(x ≥ 6) = p(x = 6) + p(x = 8) + p(x = 10)
= \(\frac { 10 }{ 36 } +\frac { 12 }{ 36 } +\frac { 9 }{ 36 } =\frac { 31 }{ 36 } \)
Sample space = {4 childrens}

Equation of the plane passing through the point
\(=\vec { a } =\hat { i } -2\hat { j } +4 \) .......(1)
Equation of the given plane is x + 2y - 3z = 11
\(\Rightarrow \vec { r } .(\hat { i } +2\hat { j } -3\hat { k } )=11\)
The given plane is perpendicular to the vector \(\hat { i } +2\hat { j } -3\hat { k } \)
∴ The required plane is parallel to the vector 
\(\vec { b } =\hat { i } +2\hat { j } -3\hat { k } \).......(2)
The given plane is parallel to the line 
\(\frac { x+7 }{ 3 } =\frac { y+3 }{ -1 } =\frac { z }{ 1 } \) Whose direction ratios are  3,-1,1
The required plane is parallel to the vector
 \(\vec { c } =3\hat { i } -\hat { j } +\hat { k } \)     (3)
∴  The non-parametric vector equation of the plane passing through a point (\(\vec { a } \)) and parallel to two vectors \(\vec { b } \) and \(\vec { c } \) is
\((\vec { r } -\vec { a } ).(\vec { b } \times \vec { c } )=0\)
\(\vec { b } \times \vec { c } =\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 1 & 2 & -3 \\ 3 & -1 & 1 \end{matrix} \right| \)
\(=\hat { i } (2-3)-\hat { j } (1+9)+\hat { k } (-1-6)\)
\(=-\hat { i } -10\hat { j } -7\hat { k } \)
\(\therefore (\vec { r } -(\hat { i } -2\hat { j } +4\hat { k } )).(-\hat { i } -10\hat { j } -7\hat { k } )=0\)
\([\vec { r } -(-\hat { i } -10\hat { j } -7\hat { k } )]-[(\hat { i } -2\hat { j } +4\hat { k } ).(-\hat { i } -10\hat { j } -7\hat { k } )]=0\)
\(\Rightarrow \vec { r } .(-\hat { i } -10\hat { j } -7\hat { k } )-[-1+20-28]=0\)
\(\Rightarrow \vec { r } .(-\hat { i } -10\hat { j } -7\hat { k } )\times 9=0\)
\(\Rightarrow \vec { r } .(\hat { i } +10\hat { j } +7\hat { k } )-9=0\)
\(\Rightarrow \vec { r } .(\hat { i } +10\hat { j } +7\hat { k } )=9\)
Let \(\Rightarrow \vec { r } =x\hat { i } +y\hat { j } +z\hat { k } \)
\(\therefore (x\hat { i } +y\hat { j } +z\hat { k } ).(\hat { i } +10\hat { j } +7\hat { k } )=9\)
⇒ x+10y+7z = 9 which is the required Cartesian equation of the plane.

The plane passes through the point.
\(\vec { a } =2\hat { i } +3\hat { j } +6\hat { k } \) and parallel to the lines \(\frac{x-1}{2}\)
\(=\frac { y+1 }{ 3 } =\frac { z-3 }{ 1 } and\frac { x+3 }{ 2 } =\frac { y-3 }{ -5 } =\frac { z+1 }{ -3 } \)
\(\Rightarrow \vec { b } =2\hat { i } +3\hat { j } +\hat { k }\ and\ \vec { c } =2\hat { i } -5\hat { j } -3\hat { k } \)
\(\vec { b } \times \vec { c } =\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 2 & 3 & 1 \\ 2 & -5 & -3 \end{matrix} \right| \)
\(=\hat { i } (-9+5)-\hat { j } (-6-2)+\hat { k } (-10-6)\)
\(=-4\hat { i } +8\hat { j } -16\hat { k } \)
The non-parametric vector equation of the plane is
\((\vec { r } .\vec { a } ).(\vec { b } \times \vec { c } )=0,\)
\(\Rightarrow [\vec { r } (2\hat { i } +3\hat { j } +16\hat { k } ).(-4\hat { i } +8\hat { j } -16\hat { k } )]=0\)
\(\Rightarrow [\vec { r } .(-4\hat { i } +8\hat { j } -16\hat { k } )]-(-8+24-96)=0\)
\(\Rightarrow \vec { r } .(-4\hat { i } +8\hat { j } -16\hat { k } )=-80\)
\(\div -4,\) We get
\(\vec { r } .(\hat { i } -2\hat { j } +4\hat { k } )=20\)
\(Let\quad \vec { r } =x\hat { i } +y\hat { j } +z\hat { k } \)
\(\Rightarrow (x\hat { i } +y\hat { j } +z\hat { k } ).(\hat { i } -2\hat { j } +4\hat { k } )=20\)
\(\Rightarrow x=2y+4z=20\)
\(\Rightarrow x-2y+4z-20=0\)

Consider a triangle ABC in which the two altitudes AD and BE intersect at O. Let CO be produced to meet AB at F. We take O as the origin and let \(\vec { OA } =\vec { a } \), \(\vec { OB } =\vec { b} \) and  \(\vec { OC } =\vec { c } \)

Since \(\vec { AD } \) is perpendicular to \(\vec { BC } \), we have \(\vec { OA } \) is perpendicular to \(\vec { BC } \), and
hence we get \(\vec { OA } \) . \(\vec { BC } \) = 0. That is, \(\vec { a } .(\vec { c } -\vec { b } )=0\), which means
\(\vec { a } .\hat{c}-\hat{a}.\hat{b}=0\)....(1)
Similarly, since \(\vec { BE } \) is perpendicular to \(\vec { CA } \), we have \(\vec { OB } \) is perpendicular to \(\vec { CA } \), and hence we get \(\vec { OB } .\vec { CA } \) = 0.
That is, \(\vec {b } .(\vec {a } -\vec { c } )=0\)
\(\vec { a } .\hat{c}-\hat{b}.\hat{c}=0\).......(2)
Adding equations (1) and (2), gives  \(\vec { a } .\hat{c}-\hat{b}.\hat{c}=0\). That is, \(\hat{c}(\hat{a}-\hat{b})=0\)
That is \(\vec { OC } \) . \(\vec { BA } \) = 0.
Therefore, \(\vec { BA } \) is perpendicular to \(\vec { OC} \). 
Which implies that \(\vec { CF} \) is perpendicular to \(\vec { AB } \). 
Hence, the perpendicular drawn from C to the side AB passes through O. Therefore, the altitudes are concurrent.

Let \(\hat { a } =\vec { OA } \) and \(\hat { b } =\vec { OB } \) be the unit vectors making angles α and β respectively, with positive x -axis, where A and B are as shown in the diagram.
Then, we get
\(\hat { a } \) = cosα\(​​\hat{i}\)+ sinα\(\hat {j } \) and \(\hat{b}\)ˆ = cosβ\(​​\hat{i}\)+ sin β \(\hat {j } \),

The angle between \(\hat { a } \) and \(\hat{b}\) is α −β and, the vectors \(\hat{b}\), \(\hat { a } \), \(\hat{k}\) form a right-handed system.
Hence, we get \(\hat { b } \times \hat { a } =\left| \hat { b } \right| \left| \hat { a } \right| \)sin(α −β )\(\hat {k}\) = sin(α −β)\(\hat {k}\)
On the other hand, \(\hat { b } \times \hat { a } =\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ cos\beta & sin\beta & 0 \\ sin\alpha & sin\alpha & 0 \end{matrix} \right| \) = (sin α cos β -cos α sin β)\(\hat { k } \)
Hence, by equations (1) and (2), we get
sin(α −β ) = sin α cos β − cos α sin β

p	q	q ➝ p	~p	~q	~q ➝ ~p
T	T	T	F	F	T
T	F	T	F	T	T
F	T	F	T	F	F
F	F	T	T	T	T

The entries in the columns corresponding q ➝ p and ~p ➝ ~q are identical and hence they are equivalent.
q ➝ p ≡ ~p ➝ ~q
Hence proved

Truth Table for ( p V q) ∧ ~q

p	q	p V q	~q	( p V q) ∧ ~q
T	T	T	F	T
T	F	T	T	T
F	T	T	F	T
F	F	F	T	T

Given u (x, y) = \(\frac { { x }^{ 2 }+{ y }^{ 2 } }{ \sqrt { x+y } } \)
\(u({ \lambda }x,{ \lambda }y)=\frac { { \lambda }^{ 2 }{ x }^{ 2 }+{ { \lambda } }^{ 2 }{ y }^{ 2 } }{ \sqrt { { \lambda }x+{ \lambda }y } } \)
= \(\frac { { { \lambda } }^{ 2 }({ x }^{ 2 }+{ y }^{ 2 }) }{ \sqrt { { \lambda } } (\sqrt { x+y } ) } \)
= \({ { \lambda } }^{ 2-\frac { 1 }{ 2 } }u(x,y)\)
= \({ { \lambda } }^{ \frac { 3 }{ 2 } }u(x,y)\)
∴ u (x, y) is a homogeneous function of degree \(\frac32\)
∴ By Euler's theorem,
\(x\frac { \partial u }{ \partial x } +y\frac { \partial u }{ \partial y } \) = n.u ≍ \(x\frac { \partial u }{ \partial x } +y\frac { \partial u }{ \partial y } =\frac { 3 }{ 2 } u\)
Hence, proved.

Let X be the random variable denotes the amount win in the lottery
Values of the random variable -2   48   98 198  Total No. of elements in inverse images 589  6 4 1 600 Probability mass function
x -2 48 98 198 f(x) \(\\ \cfrac { 589 }{ 600 } \) \(\\ \cfrac { 6 }{ 600 } \) \(\\ \cfrac { 4}{ 600 } \) \(\\ \cfrac { 1}{ 600 } \) \(E(X)=-2\times \cfrac { 589 }{ 600 } +48\times \cfrac { 6 }{ 600 } +98\times \cfrac { 4 }{ 600 } +198\times \cfrac { 1 }{ 600 } \) 
= - 0.5
Loss = Rs. 0.50

Given (x, y, z) = log (x³ + y³ + z³)
\(\frac { \partial U }{ \partial x } =\frac { 1 }{ { x }^{ 3 }+{ y }^{ 3 }+{ z }^{ 3 } } { (3x }^{ 2 });\)
\(\frac { \partial U }{ \partial y } =\frac { { 3y }^{ 2 } }{ { x }^{ 3 }+{ y }^{ 3 }+{ z }^{ 3 } } \) and
\(\frac { \partial U }{ \partial z } =\frac { { 3z }^{ 2 } }{ { x }^{ 3 }+{ y }^{ 3 }+{ z }^{ 3 } } \)
\(\therefore \frac { \partial U }{ \partial x } +\frac { \partial U }{ \partial y } +\frac { \partial U }{ \partial z } =\frac { { 3x }^{ 2 } }{ { x }^{ 3 }+{ y }^{ 3 }+{ z }^{ 3 } } +\frac { { 3y }^{ 2 } }{ { x }^{ 3 }+{ y }^{ 3 }+{ z }^{ 3 } } +\frac { { 3z }^{ 2 } }{ { x }^{ 3 }+{ y }^{ 3 }+{ z }^{ 3 } } \)
\(=\frac { { 3({ x }^{ 2 }+y }^{ 2 }+{ z }^{ 2 }) }{ { x }^{ 3 }+{ y }^{ 3 }+{ z }^{ 3 } } \)

Given \(f(x)=\begin{cases} \begin{matrix} kxe^{ -2x } & forx>0 \end{matrix} \\ \begin{matrix} 0 & for\quad x\le 0 \end{matrix} \end{cases}\)
Since the given function is a probability density function 
\(\int _{ -\infty }^{ \infty }{ f(x)dx } \) = 1 
\(\Rightarrow k\int _{ 0 }^{ \infty }{ { xe }^{ -2x }dx=1 } \)
\(\Rightarrow k \frac { 1! }{ \left( 2 \right) ^{ 2 } } =1\) 
[\(\int _{ 0 }^{ \infty }{ { x }^{ n }e^{ -ax } } =\frac { n! }{ { a }^{ +1 } } \), Here a = 2, n = 1]
\(\Rightarrow \frac { k }{ 4 } =1\\ \Rightarrow k=4\)

Let X be the random variable denotes the total C score is two throws of a die.
Sample space S

II	2	3	3	4	4	4
I
2	4	5	5	6	6	6
3	5	6	6	7	7	7
3	5	6	6	7	7	7
4	6	7	7	8	8	8
4	6	7	7	8	8	8
4	6	7	7	8	8	8

n (S) = 36
X = {4,5,6,7,8}
From the sample space

Values of random variable	4	5	6	7	8	Total
No of points in inverse image	1	4	10	12	9	36

\(\vec { a } =\hat { i } +2\hat { j } +3\hat { k } ,\vec { b } =2\hat { i } -\hat { j } +\hat { k } ,\vec { c } =3\hat { i } +2\hat { j } +\hat { k } \)
\(\vec { a } .\vec { c } =(\hat { i } +2\hat { j } +3\hat { k } ),(3\hat { i } +2\hat { j } +\hat { k } )\)
= 3+4+3 = 10
\(\vec { a } .\vec { b } =(\hat { i } +2\hat { j } +3\hat { k } ).(2\hat { i } -\hat { j } +\hat { k } )\)
= 2 - 2 + 3 = 3
∴ \((\vec { a } .\vec { c } )\vec { b } -(\vec { a } .\vec { b } )\vec { c } =10\vec { b } -3\vec { c } \)
⇒ \(\vec { a } \times (\vec { b } \times \vec { c } )=10\vec { b } -3\vec { c } \) ..(1)
\([\vec { a } \times (\vec { b } \times \vec { c } )=(\vec { a } .\vec { c } )\vec { b } -(\vec { a } .\vec { b } )\vec { c } ]\)
Given \(\vec { a } \times (\vec { b } \times \vec { c } )=l\vec { a } +m\vec { b } +n\vec { c } \)..(2)
From (1) & (2)
\(10\vec { b } -3\vec { c } =l\vec { a } +m\vec { b } +n\vec { c } \)
Comparing the co-efficients of like terms, we get
l = 0, m = 10 and n = -3

Given \(\vec { a } =-2\hat { i } +5\hat { j } +3\hat { k } \), \(\hat { b } =\hat { i } +3\hat { j } -2\hat { k } \) and \(\vec { c } =-3\vec { i } +\vec { j } +4\vec { k } \)
Volume of the parallelepiped = \(\vec { a } .(\vec { b } \times \vec { c } )\)
= \(\left| \begin{matrix} -2 & 5 & 3 \\ 1 & 3 & -2 \\ -3 & 1 & 4 \end{matrix} \right| \)
= \(-2\left| \begin{matrix} 3 & -2 \\ 1 & 4 \end{matrix} \right| -5\left| \begin{matrix} 1 & -2 \\ - & 4 \end{matrix} \right| +3\left| \begin{matrix} 1 & 3 \\ -3 & 1 \end{matrix} \right| \)
= -2(12+2)-5(4-6)+3(1+9)
= -2(14)-5(-2)+3(10) = -28+10+30
= 12
Vector Area of the base parallelogram = \(\vec { b } \times \vec { c } \)
=\(\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 1 & 3 & -2 \\ -3 & 1 & 4 \end{matrix} \right| =\hat { i } \left| \begin{matrix} 3 & -2 \\ 1 & 4 \end{matrix} \right| \hat { j } \left| \begin{matrix} 1 & -2 \\ -3 & 4 \end{matrix} \right| +\hat { k } \left| \begin{matrix} 1 & 3 \\ -3 & 1 \end{matrix} \right| \)
= \(\hat { i } \)(12+2)-\(\hat { j } \)(4-6)+\(\hat { k } \)(1+9)
= \(\hat { i } \)(14)-\(\hat { j } \)(-2)+\(\hat { k } \)(10)
= 14\(\hat { k } \)+2\(\hat { j } \)+10\(\hat { k } \)
Area of the parallelogram =\(\sqrt { { 14 }^{ 2 }+{ 2 }^{ 2 }+{ 10 }^{ 2 } } \)
= \(\sqrt { 196+4+100 } \)
= \(\\ \sqrt { 300 } =10\sqrt { 3 } \)..............(2)
Volume of the parallelepiped = Base area x attitude
∴ Altitude = \(\frac { Volume }{ Base\quad area } =\frac { 12 }{ 10\sqrt { 3 } } =\frac { 6 }{ 5\sqrt { 3 } } \times \frac { \sqrt { 3 } }{ \sqrt { 3 } } \)

= \(\frac { 2\sqrt { 3 } }{ 5 } \) units.

Given \(\vec { F } =3\hat { i } +4\hat { j } -5\hat { k } \)
\(\vec { r } \) = (Force acting through the point) - (force acting to the point)
= \((4\hat { i } +2\hat { j } -3\hat { k } )-(2\hat { i } -\hat { j } +4\hat { k } )\)
= \(2\hat { i } +5\hat { j } -7\hat { k } \)
Torque = \(\vec { c } =\hat { r } \times \hat { F } \)
=  \(\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 2 & 5 & -7 \\ 3 & 4 & -5 \end{matrix} \right| =\hat { i } \left| \begin{matrix} 5 & -7 \\ 4 & -5 \end{matrix} \right| -\hat { j } \left| \begin{matrix} 2 & -7 \\ 3 & -5 \end{matrix} \right| +\hat { k } \left| \begin{matrix} 2 & 5 \\ 3 & 4 \end{matrix} \right| \)
\(\hat { i } =\hat { i } (-25+28)-\hat { j } (-10+21)+\hat { j } (8-15)\)
= \(3\hat { i } -11\hat { j } -7\hat { k } \)
∴ MagnitudeoftheTorque = \(\sqrt { { 3 }^{ 2 }+(-11)^{ 2 }+(-7)^{ 2 } } \)
= \(\sqrt { 9+121+49 } =\sqrt { 179 } \)
Hence, the direction cosines are \(\left( \frac { 3 }{ \sqrt { 179 } } ,\frac { -11 }{ \sqrt { 179 } } ,\frac { -7 }{ \sqrt { 179 } } \right) \).

Given p : Jupiter is a planet and
q : India is an island.
(i) ¬p : Jupiter is not a planet.
(ii) p ∧ ¬q : Jupiter is a planet and India is not an island.
(iii) ¬p ∨ q : Jupiter is not a planet or India is an island.
(iv) p➝ ¬q : If Jupiter is a planet then India is not an island.
(v) p↔q : Jupiter is a planet if and only if India is an island.

√b is not defined for negative values, b which also ∈ R.
Hence, a√b is not defined for all a, b ∈R
* is not a binary operation on R

\(h(x,y)=\frac { 6{ x }^{ 2 }{ y }^{ 3 }-\pi { y }^{ 5 }+9{ x }^{ 4 }y }{ 2020{ x }^{ 2 }+2019{ y }^{ 2 } } \)
Given \(h(x,y)=\frac { 6{ x }^{ 2 }{ y }^{ 3 }-\pi { y }^{ 5 }+9{ x }^{ 4 }y }{ 2020{ x }^{ 2 }+2019{ y }^{ 2 } } \)
\(h(\lambda x,\lambda y)=\frac { 6{ \lambda }^{ 2 }{ x }^{ 2 }{ \lambda }^{ 3 }{ y }^{ 3 }-\pi { \lambda }^{ 5 }{ y }^{ 5 }+9{ \lambda }^{ 4 }{ x }^{ 4 }\lambda y }{ 2020{ \lambda }^{ 2 }{ x }^{ 2 }+2019{ \lambda }^{ 2 }{ y }^{ 2 } } \)
\(=\frac { { \lambda }^{ 5 }(6{ x }^{ 2 }{ y }^{ 3 }-\pi { y }^{ 5 }+9{ x }^{ 4 }y) }{ { \lambda }^{ 2 }(2020{ x }^{ 2 }+2019{ y }^{ 2 }) } \)
\( =\lambda^{3} \mathrm{~h}(\mathrm{x}, \mathrm{y}) \)
Thus f is homogeneous with degree 3.

Given f(x, y) = x²y + 6x³ + 7
f(λx, λy) = λ²x²λy + 6λ³x³ + 7
= λ³ x² y + 6λ³ x³ + 7
≠ λf(x, y)
There is no common λ in this equation.
It is not homogeneous

Given n = 6, \(p=\frac { 1 }{ 3 } \), k = 3
\(P(X=k)=\left( \begin{matrix} n \\ k \end{matrix} \right) { p }^{ k }\left( 1-p \right) ^{ n-k },\) 
n = 0,1,2, ... n
\(\therefore P(X=k)=\left( \begin{matrix} n \\ k \end{matrix} \right) { p }^{ k }(1-p)^{ n-k }\) 
n = 0,1,2, ... n
\(P(X=3)=\left( \begin{matrix} 6 \\ 3 \end{matrix} \right) \left( \cfrac { 1 }{ 3 } \right) ^{ 3 }\left( 1-p \right) ^{ 6-3 }\) 
= \(\left( \begin{matrix} 6 \\ 3 \end{matrix} \right) \left( \cfrac { 1 }{ 3 } \right) ^{ 3 }\left( \cfrac { 2 }{ 3 } \right) ^{ 2 }\) 

\(P(X=3)=\frac { 160 }{ 729 } \) 

\(f(x)=\begin{cases} \begin{matrix} \frac { 1 }{ 2 } e^{ -\frac { x }{ 2 } } & for\quad x>0 \end{matrix} \\ \begin{matrix} 0 & otherwise \end{matrix} \end{cases}\)
\(\int _{ 0 }^{ \infty }{ x.f(x)dx } =\frac { 1 }{ 2 } \int _{ 0 }^{ \infty }{ x.{ e }^{ \frac { -x }{ 2 } } } dx\) 
\(\left[ \int _{ 0 }^{ \infty }{ { e }^{ -ax }.{ x }^{ n }dx=\cfrac { n! }{ { a }^{ n+1 } } } \right] \) 
= \(\frac { 1 }{ 2 } \times \frac { 1! }{ \left( \frac { 1 }{ 2 } \right) ^{ 2 } } =\frac { 1 }{ 2 } \times \frac { 1 }{ \frac { 1 }{ 4 } } \) 
= \(\frac { 1 }{ 2 } \times \frac { 4 }{ 1 } =2\) 
\(E({ X }^{ 2 })=\int _{ 0 }^{ \infty }{ { x }^{ 2 }.f(x) } dx\) 
= \(\int _{ 0 }^{ \infty }{ { x }^{ 2 }.\frac { 1 }{ 2 } { e }^{ -\frac { x }{ 2 } } } dx\) 
= \(\frac { 1 }{ 2 } \int { { x }^{ 2 }.{ e }^{ -\frac { x }{ 2 } }dx } \) 
= \(\frac { 1 }{ 2 } \times \frac { 2! }{ \left( \frac { 1 }{ 3 } \right) ^{ 3 } } =\frac { 1 }{ 2 } \times \frac { 2 }{ \frac { 1 }{ 8 } } \) 
= \(\frac { 1 }{ 2 } \times 2\times 8=8\) 
ஃVar(X)=E(X²) - [E(x)]²
= 8-2²
= 8 - 4 = 4

x	0	1	2	3	4	5
f(x)	\(\cfrac { 1 }{ 5 } \)	\(\cfrac { 1 }{ 5 } \)	\(\cfrac { 1 }{ 10 } \)	\(\cfrac { 1 }{ 5 } \)	\(\cfrac { 1 }{ 5 } \)	\(\cfrac { 1 }{ 10 } \)

\(\therefore Mean=E(x)=\Sigma xf(x)=0\left( \frac { 1 }{ 5 } \right) +1\left( \frac { 1 }{ 5 } \right) +2\left( \frac { 1 }{ 10 } \right) +3\left( \frac { 1 }{ 5 } \right) +4\left( \frac { 1 }{ 5 } \right) +5\left( \frac { 1 }{ 10 } \right) \) 
\(\frac { 1 }{ 5 } +\frac { 1 }{ 5 } +\frac { 3 }{ 5 } +\frac { 4 }{ 5 } +\frac { 1 }{ 2 } \) 
= \(\frac { 2+2+6+8+5+ }{ 10 } =\frac { 23 }{ 10 } =2.3\) 
= \(f({ x }^{ 2 })=\Sigma { x }^{ 2 }f(x)\)
= \({ 0 }^{ 2 }\left( \frac { 1 }{ 5 } \right) +{ 1 }^{ 2 }\left( \frac { 1 }{ 5 } \right) +2^{ 2 }\left( \frac { 1 }{ 10 } \right) +{ 3 }^{ 2 }\left( \frac { 1 }{ 5 } \right) +{ 4 }^{ 2 }\left( \frac { 1 }{ 5 } \right) +{ 5 }^{ 2 }\left( \frac { 1 }{ 10 } \right) \) 
= \(0+\frac { 1 }{ 5 } +\frac { 4 }{ 10 } +\frac { 9 }{ 5 } +\frac { 16 }{ 5 } +\frac { 25 }{ 10 } \) 
= \(\frac { 2+4+18+32+25 }{ 10 } =\frac { 81 }{ 10 } =8.1\) 
= \(\frac { 2+4+18+32+25 }{ 10 } =\frac { 81 }{ 10 } =8.1\)
Variance = E(X²) - [E(X)]²
= 8.1- (2.3)2
= 8.1- 5.29
= 2.81

The angle between a line \(\vec { r } =\vec { a } +t\vec { b } \) and a plane \(\vec { r } .\vec { n } \) = p with normal \(\vec { n } \) is θ = \(sin^{ -1 }\left( \frac { |\vec { b } .\vec { n } | }{ |\vec { b }| .|\vec { n } | } \right) \quad \)
Here, \(\vec { b } =\hat { i } -\hat { j } +\hat { k } \) and \(\vec { n } =2\hat { i } \hat { j } +\hat { k } \)
So, we get θ = \(sin^{ -1 }\left( \frac { |\vec { b } .\vec { n } | }{ |\vec { b }| .|\vec { n } | } \right) \quad =sin^{ -1 }\left( \frac { |(\hat { i } -\hat { j } +\hat { k } ).(2\hat { i } -\hat { j } +\hat { k } )| }{ |\hat { i } -\hat { j } +\hat { k } ||2\hat { i } -\hat { j } +\hat { k } | } \right) =sin^{ -1 }\left( \frac { 2\sqrt { 3 } }{ 3 } \right) \).

  \(\vec { a } \)= \(a\hat { i } +a\hat { j } +c\hat { k } , \vec{b}=\hat { i } +\hat { k } \), \(\vec { c } \)= \(c\hat { i } +c\hat { j } +b\hat { k } \)
Given \(\vec { a } ,\vec { b } \) and \(\vec { c } \) are co-planar
\(\vec { a } .(\vec { b } \times \vec { c } )\) = 0
⇒ \(\left| \begin{matrix} a & a & c \\ 1 & 0 & 1 \\ c & c & b \end{matrix} \right| \) = 0
⇒ \(a\left| \begin{matrix} 0 & 1 \\ c & b \end{matrix} \right| -a\left| \begin{matrix} 1 & 1 \\ c & b \end{matrix} \right| +c\left| \begin{matrix} 1 & 0 \\ c & c \end{matrix} \right| \) = 0
⇒ a(0-c)-a(b-c)+c(c-0) = 0
\(\Rightarrow-\not a c-a b+\not a c+c^{2}=0\)
⇒ c² = ab ⇒ c =\(\sqrt { ab } \).
Hence c is the geometric mean of a and b.

Given \(\vec { a } =\hat { i } -2\hat { j } +3\hat { k }, \vec { b } =2\hat { i } +\hat { j } -2\hat { k }, \vec { c } =3\hat { i } +2\hat { j } +\hat { k } \)
∴ \(\vec { a } .(\vec { b } \times \vec { c } )=\left| \begin{matrix} 1 & -2 & 3 \\ 2 & 1 & -2 \\ 3 & 2 & 1 \end{matrix} \right| \)
= \(1\left| \begin{matrix} 1 & -2 \\ 2 & 1 \end{matrix} \right| +2\left| \begin{matrix} 2 & -2 \\ 3 & 1 \end{matrix} \right| +3\left| \begin{matrix} 2 & 1 \\ 3 & 2 \end{matrix} \right| \)
= 1(1+4)+2(2+6)+3(4-3)
= 1(5)+2(8)+3(1)
= 5+16+3 = 24
\(\vec { a } .(\vec { b } \times \vec { c } )\) = 24