Vector Algebra

$ 4 \hat{\mathbf{i}}+\hat{\mathbf{j}}+3 \hat{\mathbf{k}}, 6 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}-3 \hat{\mathbf{k}}, \hat{\mathbf{i}}-\hat{\mathbf{j}}-3 \hat{\mathbf{k}} $

$\Rightarrow$ points are $P(4,1,3), Q(6,-2,-3)$ and $R(1,-1,-3)$

$ \begin{aligned} &\begin{aligned} \Rightarrow P Q & =\sqrt{(6-4)^2+(-3)^2+(-6)^2} \\ & =\sqrt{4+9+36}=\sqrt{49}=7 \\ \Rightarrow Q R & =\sqrt{(1-6)^2+(-1+2)^2+(-3+3)^2} \\ & =\sqrt{(-5)^2+(1)^2+0}=\sqrt{26} \\ \Rightarrow P R & =\sqrt{(1-4)^2+(-1-1)^2+(-3-3)^2} \\ & =\sqrt{9+4+36}=\sqrt{49}=7 \\ \Rightarrow P Q & =P R=7 \text { units } \end{aligned}\\ &\Rightarrow \triangle P Q R \text { is an isosceles triangle. } \end{aligned} $

$ \begin{aligned} &\text { Magnitude of the vector }-2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}}\\ &\begin{aligned} & \Rightarrow|\mathbf{b}|=\sqrt{4+1+4}=\sqrt{9}=3 \\ & \Rightarrow \text { Unit vector } \hat{\mathbf{b}}=\frac{-2 \hat{\mathbf{i}}}{3}-\frac{1}{3} \hat{\mathbf{j}}+\frac{2}{3} \hat{\mathbf{k}} \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { Now, let } \mathbf{c}=\hat{\mathbf{i}}-7 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}\\ &\begin{aligned} & |\mathbf{c}|=\sqrt{54}=3 \sqrt{6} \\ & \hat{\mathbf{c}}=\frac{1}{3 \sqrt{6}} \hat{\mathbf{i}}-\frac{7}{3 \sqrt{6}} \hat{\mathbf{j}}+\frac{2}{3 \sqrt{6}} \hat{\mathbf{k}} \\ & \Rightarrow \mathbf{c}=\mathbf{a}+\mathbf{b} \Rightarrow \mathbf{a}=\mathbf{c}-\mathbf{b} \\ & =\frac{(1+2 \sqrt{6}) \hat{\mathbf{i}}}{3 \sqrt{6}}+\frac{(-7+\sqrt{6}) \hat{\mathbf{j}}}{3 \sqrt{6}}+\frac{(2-2 \sqrt{6}) \hat{\mathbf{k}}}{3 \sqrt{6}} \\ & \text { Compaire with } \mathbf{a}=x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}} \\ & \Rightarrow x=\frac{1+2 \sqrt{6}}{3 \sqrt{6}}=\frac{\sqrt{6}+12}{18} \cong \frac{7}{9} \\ & \Rightarrow x=\frac{7}{9} \end{aligned} \end{aligned} $

$ \text { } \begin{aligned} & \mathbf{a}=2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+6 \hat{\mathbf{k}}, \mathbf{b}=\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}, \\ & \mathbf{c}=3 \hat{\mathbf{j}}-\hat{\mathbf{k}} \\ & \mathbf{a} \times \mathbf{b}=\left|\begin{array}{rrr} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 2 & -1 & 6 \\ 1 & -1 & 1 \end{array}\right| \\ &=\hat{\mathbf{i}}(-1+6)-\hat{\mathbf{j}}(2-6)+\hat{\mathbf{k}}(-2+1) \\ &=5 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}-\hat{\mathbf{k}} \\ & \mathbf{b} \times \mathbf{c}=\left|\begin{array}{rrr} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & -1 & 1 \\ 0 & 3 & -1 \end{array}\right| \\ &=\hat{\mathbf{i}}(1-3)-\hat{\mathbf{j}}(-1-0)+\hat{\mathbf{k}}(3-0) \\ &=-2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+3 \hat{\mathbf{k}} \end{aligned} $

$ \begin{aligned} \mathbf{c} \times \mathbf{a} & =\left|\begin{array}{rrr} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 0 & 3 & -1 \\ 2 & -1 & 6 \end{array}\right| \\ & =\hat{\mathbf{i}}(18-1)-\hat{\mathbf{j}}(+2)+\hat{\mathbf{k}}(-6) \\ & =17 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}-6 \hat{\mathbf{k}} \\ \mathbf{a} \times \mathbf{b} & +\mathbf{b} \times \mathbf{c}+\mathbf{c} \times \mathbf{a} \\ & =5 \hat{\mathbf{i}}+4 \hat{\mathbf{j}}-\hat{\mathbf{k}}+(-2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+3 \hat{\mathbf{k}})+17 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}-6 \hat{\mathbf{k}} \\ & =20 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-4 \hat{\mathbf{k}} \end{aligned} $

$\mathbf{a}=2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}}$ and $\mathbf{b}=\hat{\mathbf{i}}+\hat{\mathbf{j}}$

$ \mathbf{a} \cdot \mathbf{c}=|\mathbf{c}|,|\mathbf{c}-\mathbf{a}|=2 \sqrt{2} $

$\mathbf{a} \times \mathbf{b}$ and $\mathbf{c}$ is $30^{\circ}$.

$ \begin{aligned} & |(\mathbf{a} \times \mathbf{b}) \times \mathbf{c}|=|\mathbf{a} \times \mathbf{b}||\mathbf{c}| \sin \theta \\ & \quad=|\mathbf{a} \times \mathbf{b}||\mathbf{c}| \times \frac{1}{2} \\ & \mathbf{a}=2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}}, \mathbf{b}=\hat{\mathbf{i}}+\hat{\mathbf{j}} \\ & \mathbf{a} \times \mathbf{b}=\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 2 & 1 & -2 \\ 1 & 1 & 0 \end{array}\right|=\hat{\mathbf{i}}(2)-\hat{\mathbf{j}}(2)+\hat{\mathbf{k}}(1) \\ & \quad=2 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}} \\ & |\mathbf{a} \times \mathbf{b}|=\sqrt{4+4+1}=3 \\ & \quad \mathbf{a} \cdot \mathbf{c}=|\mathbf{c}| \\ & \Rightarrow|\mathbf{a} \| \mathbf{c}| \cos \alpha=|\mathbf{c}| \end{aligned} $

$ \begin{aligned} & \cos \alpha=\frac{1}{3} \\ & |\mathbf{c}-\mathbf{a}|=2 \sqrt{2} \\ & \Rightarrow|\mathbf{c}-\mathbf{a}|^2=8 \\ & \Rightarrow|\mathbf{c}|^2+|\mathbf{a}|^2-2|\mathbf{c}||\mathbf{a}| \cos \alpha=8 \\ & \Rightarrow|\mathbf{c}|^2+9-2|\mathbf{c}| 3 \times \frac{1}{3}=8 \\ & \Rightarrow|\mathbf{c}|^2-2|\mathbf{c}|+1=0 \\ & \Rightarrow(|\mathbf{c}|-1)^2=0 \Rightarrow|\mathbf{c}|=1 \\ & \begin{aligned} \mid(\mathbf{a} \times \mathbf{b}) & \times \mathbf{c}|=|\mathbf{a} \times \mathbf{b}|| \mathbf{c} \mid \sin \theta \\ & =3 \times 1 \times \sin 30^{\circ} \\ & =3 \times 1 \times \frac{1}{2}=\frac{3}{2} \end{aligned} \end{aligned} $

$ \begin{aligned} &(a+2 b-c)\{(a-b) \times(a-b-c)\} \\ &=(a+2 b-c)[a \times a-a \times b-a \times c-b\times a+b \times b+b \times c] \end{aligned} $

$ \begin{aligned} & =(\mathbf{a}+2 \mathbf{b}-\mathbf{c}) [-\mathbf{a} \times \mathbf{b}-\mathbf{a} \times \mathbf{c}-\mathbf{b} \times \mathbf{a}+\mathbf{b} \times \mathbf{c}] \\ & =(\mathbf{a}+2 \mathbf{b}-\mathbf{c})[\mathbf{b} \times \mathbf{a}+\mathbf{c} \times \mathbf{a}+\mathbf{a} \times \mathbf{b}+\mathbf{b} \times \mathbf{c}] \end{aligned} $

$ \begin{aligned} = & \mathbf{a} \cdot(\mathbf{b} \times \mathbf{a})+\mathbf{a} \cdot(\mathbf{c} \times \mathbf{a})+\mathbf{a} \cdot(\mathbf{a} \times \mathbf{b}) \\ & +\mathbf{a} \cdot(\mathbf{b} \times \mathbf{c}) \\ & +2 \mathbf{b} \cdot(\mathbf{b} \times \mathbf{a})+2 \mathbf{b} \cdot(\mathbf{c} \times \mathbf{a})+2 \mathbf{b}(\mathbf{a} \times \mathbf{b}) \\ & +2 \mathbf{b} \cdot(\mathbf{b} \times \mathbf{c}) \\ & -\mathbf{c} \cdot(\mathbf{b} \times \mathbf{a})-\mathbf{c}(\mathbf{c} \times \mathbf{a})-\mathbf{c}(\mathbf{a} \times \mathbf{b})-\mathbf{c}(\mathbf{b} \times \mathbf{c}) \\ = & {[\mathbf{a b c}]+2[\mathbf{b c a}]-[\mathbf{c b a}]-[\mathbf{c a b}] } \\ = & {[\mathbf{a b c}]+2[\mathbf{a b c}]+[\mathbf{a b c}]-[\mathbf{a b c}]=3[\mathbf{a b c}] } \end{aligned} $

Given that

$ \begin{aligned} \mathrm{OP} & =\hat{\mathbf{i}}-\hat{\mathbf{j}}-\hat{\mathbf{k}}, \mathrm{OQ}=-\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}} \\ \mathbf{a} & =\hat{\mathbf{i}}+\hat{\mathbf{j}}, \mathbf{b}=\hat{\mathbf{j}}-\hat{\mathbf{k}} \\ l_1 & =\mathbf{r}_1=\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}+\lambda(\hat{\mathbf{i}}+\hat{\mathbf{j}}) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ l_2 & =\mathbf{r}_2=-\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}+\mu(\hat{\mathbf{j}}-\hat{\mathbf{k}}) \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{aligned} $

Let point of intersection of first line.

$ \begin{aligned} &\begin{aligned} \mathbf{L} & =\mathbf{P}+\lambda \mathbf{a}=(1+\lambda) \hat{\mathbf{i}}+(-1+\lambda) \hat{\mathbf{j}}-\hat{\mathbf{k}} \\ M & =\mathbf{Q}+\mu \mathbf{b}=-\hat{\mathbf{i}}+(1+\mu) \hat{\mathbf{j}}+(1-\mu) \hat{k} \end{aligned}\\ &\text { Line passing through } L \text { and } M\\ &\begin{array}{lc} \therefore \mathbf{L M} \| \mathbf{C} \\ \mathbf{L M}=(-2-\lambda) \hat{\mathbf{i}}+(2+\mu-\lambda) \hat{\mathbf{j}}+(2-\mu) \hat{k} \end{array} \end{aligned} $

$ \begin{aligned} & \mathbf{L} \mathbf{M}=t(\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}) \\ & -2-\lambda=t \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\\ & \Rightarrow \quad \lambda=-2-t \\ & 2+\mu-\lambda=-t \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iv)\\ & 2-\mu=t \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(v)\\ & \Rightarrow \quad \mu=2-t \\ & t=-6, \lambda=4 \\ & \mu=8 \\ & \mathbf{M}=-\hat{\mathbf{i}}+9 \hat{\mathbf{j}}-7 \hat{\mathbf{k}} \\ & \mathbf{P M}=-2 \hat{\mathbf{i}}+10 \hat{\mathbf{j}}-6 \hat{\mathbf{k}} \end{aligned} $

$ \begin{aligned} & \mathbf{p}=(a+1) \hat{\mathbf{i}}+a \hat{\mathbf{j}}+a \hat{\mathbf{k}} \\ & \mathbf{q}=a \hat{\mathbf{i}}+(a+1) \hat{\mathbf{j}}+a \hat{\mathbf{k}} \\ & \mathbf{r}=a \hat{\mathbf{i}}+a \hat{\mathbf{j}}+(a+1) \hat{\mathbf{k}} \end{aligned} $

$\mathbf{p}, \mathbf{q}$ and $\mathbf{r}$ are coplanar

$ \begin{aligned} & \left|\begin{array}{ccc} a+1 & a & a \\ a & a+1 & a \\ a & a & a+1 \end{array}\right|=0 \\ \Rightarrow & (3 a+1)\left|\begin{array}{ccc} 1 & 1 & 1 \\ a & a+1 & a \\ a & a & a+1 \end{array}\right|=0 \\ \Rightarrow & (3 a+1)\left[(a+1)^2-a^2-1\left(a^2+a-a^2\right)+1\right. \\ & \left.\left(a^2-a^2-a\right)\right]=0 \end{aligned} $

$ \begin{aligned} & \Rightarrow(3 a+1)[2 a+1-a-a] \Rightarrow a=-\frac{1}{3} \\ & \mathbf{p} \cdot \mathbf{q}=-\frac{2}{9}-\frac{2}{9}+\frac{1}{9}=\frac{-3}{9}=\frac{-1}{3} \\ & \begin{aligned} & \Rightarrow|\mathbf{r} \times \mathbf{q}|=\sqrt{\left(\frac{4}{9}-\frac{1}{9}\right)^2-\left(\frac{-2}{9}-\frac{1}{9}\right)^2} +\left(\frac{1}{9}+\frac{2}{9}\right)^2 \end{aligned} \\ & \\ & =\sqrt{\frac{9-9+9}{81}}=\frac{1}{3} \\ & 3(\mathbf{p} \cdot \mathbf{q})^2-\lambda|\mathbf{r} \times \mathbf{q}|^2=0 \Rightarrow \lambda=1 \end{aligned} $

$ \text { } \begin{aligned} \mathbf{a}= & \hat{\mathbf{i}}+4 \hat{\mathbf{j}}-4 \hat{\mathbf{k}} \\ & \mathbf{b}=-2 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}-2 \hat{\mathbf{k}} \\ & \mathbf{c}=3 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}-4 \hat{\mathbf{k}} \\ \Rightarrow & (\mathbf{b} \times \mathbf{c}) \times \mathbf{a}=x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}} \\ \Rightarrow & (\mathbf{b} \cdot \mathbf{a}) \mathbf{c}-(\mathbf{c} \cdot \mathbf{a}) \mathbf{b}=x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}} \\ \Rightarrow & (-2+20+8) \mathbf{c}-(3-8+16) \mathbf{b} \\ & =x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}} \\ \Rightarrow & 100 \hat{\mathbf{i}}+107 \hat{\mathbf{j}}-82 \hat{\mathbf{k}}=x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}} \\ \Rightarrow & x+y-z=100-107+82=75 \end{aligned} $

$ \begin{aligned} & \mathbf{A B}=5 \hat{\mathbf{i}}-4 \hat{\mathbf{j}}+15 \hat{\mathbf{k}} \\ & \mathbf{A C}=7 \hat{\mathbf{i}}+20 \hat{\mathbf{j}}+3 \hat{\mathbf{k}} \end{aligned} $

$ \begin{aligned} \cos \theta & =\frac{35-80+45}{|\mathbf{A B}||\mathbf{A C}|}=0 \\ \theta & =90^{\circ} \end{aligned} $

$ \text { We have, } \mathbf{x} \cdot(\mathbf{c}-\mathbf{b})=\mathbf{a} \cdot \mathbf{c}-\mathbf{a} \cdot \mathbf{b} $

$ \begin{aligned} & \quad=\mathbf{a}(\mathbf{c}-\mathbf{b}) \\ & (\mathbf{x}-\mathbf{a}) \cdot(\mathbf{c}-\mathbf{b})=0 \\ & \quad \mathbf{A P} \cdot \mathbf{B C}=0 \\ & \quad \mathbf{A P} \perp \mathbf{B C} \\ & \text { and } \mathbf{x} \cdot(\mathbf{a}-\mathbf{c})=\mathbf{a} \cdot \mathbf{b}-\mathbf{b} \cdot \mathbf{c}=(\mathbf{a}-\mathbf{c}) \mathbf{b} \\ & \Rightarrow(\mathbf{x}-\mathbf{b}) \cdot(\mathbf{a}-\mathbf{c})=0 \\ & \quad \mathbf{B P} \cdot \mathbf{A C}=0 \\ & \therefore \quad \mathbf{B P} \perp \mathbf{A C} \end{aligned} $

$\because$ Point of intersection of altitude is called orthocentre.

$\because P$ is orthocentre of $\triangle A B C$.

Given $|\mathbf{a}|=2,|\mathbf{b}|=3,|\mathbf{c}|=5$

$ \begin{aligned} & (a \cdot b)=\frac{\pi}{3} \\ & a \cdot b=3 \text { and } b \cdot c=\frac{15}{2} \end{aligned} $

Given, $|\mathbf{a}+\mathbf{b}+\mathbf{c}|=\sqrt{69}$

On squaring both sides, we get

$ \begin{aligned} & |\mathbf{a}|^2+|\mathbf{b}|^2+|\mathbf{c}|^2+2(\mathbf{a} \cdot \mathbf{b}+\mathbf{b} \cdot \mathbf{c}+\mathbf{c} \cdot \mathbf{a}) =69\\ & \Rightarrow \quad 4+9+25+2\left(3+\frac{15}{2}+\mathbf{c} \cdot \mathbf{a}\right)=69 \\ & \Rightarrow 2\left(\frac{21}{2}+\mathbf{c} \cdot \mathbf{a}\right)=31 \\ & \Rightarrow \quad 21+2 \cdot \mathbf{c}=31 \\ & \Rightarrow \quad 2(\mathbf{c} \cdot \mathbf{a})=10 \\ & \Rightarrow \quad \mathbf{c} \cdot \mathbf{a}=5 \\ & \Rightarrow \quad|\mathbf{c}||\mathbf{a}| \cos \theta=5 \\ & \Rightarrow \quad 5 \times 2 \cos \theta=5 \\ & \Rightarrow \quad \cos \theta=\frac{1}{2} \Rightarrow \theta=\frac{\pi}{3} \end{aligned} $

$ \begin{aligned} &\text { We have, } \mathbf{O A}=\hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}} \text {, }\\ &\begin{aligned} & \mathrm{OB}=\hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}} \\ & \mathrm{OC}=\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}} \end{aligned} \end{aligned} $

$ \mathrm{OD}=2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}} $

Let $\mathbf{n}_1$ is the vector normal to the face $A B C$

$ \because \mathbf{n}_1=\mathbf{A B} \times \mathbf{A C} $

$ \begin{aligned} \mathbf{n}_1 & =\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 0 & -2 & 3 \\ 0 & -3 & 2 \end{array}\right| \\ & =\hat{\mathbf{i}}(-4+9)+\hat{\mathbf{j}}(0)+\hat{\mathbf{k}}(0)=5 \hat{\mathbf{i}} \end{aligned} $

Let $\mathbf{n}_2$ be the vector normal to face $A B D$

$ \because \mathbf{n}_2=\mathbf{A B} \times \mathbf{A D} $

$ n_2=\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 0 & -2 & 3 \\ 1 & 0 & 2 \end{array}\right|=\hat{\mathbf{i}}(-4)-\hat{\mathbf{j}}(-3)+\hat{\mathbf{k}}(2) $

$ \mathbf{n}_2=-4 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+2 \hat{\mathbf{k}} $

Now, angle between $\mathbf{n}_1$ and $\mathbf{n}_2$

$ \begin{aligned} \cos \theta & =\frac{\mathbf{n}_1 \cdot \mathbf{n}_2}{\left|\mathbf{n}_1\right|\left|\mathbf{n}_2\right|}=\frac{-20}{5 \sqrt{29}} \\ \theta & =\cos ^{-1}\left(\frac{-4}{\sqrt{29}}\right) \end{aligned} $

Given, $|\mathbf{a}|=|\mathbf{b}|=|\mathbf{c}|=1$

$ \begin{aligned} & \mathbf{a} \cdot \mathbf{b}=0 \\ & (\mathbf{a}-\mathbf{c}) \cdot(\mathbf{b}+\mathbf{c})=0 \\ & \mathbf{a} \cdot \mathbf{b}+\mathbf{a} \cdot \mathbf{c}-\mathbf{c} \cdot \mathbf{b}-|\mathbf{c}|^2=0 \\ & \mathbf{a} \cdot \mathbf{c}-\mathbf{b} \cdot \mathbf{c}=1 \\ & \text { Also, } \mathbf{c}=l \mathbf{c}+m \mathbf{b}+n(\mathbf{a} \times \mathbf{b}) \\ & \mathbf{a} \cdot \mathbf{c}=l(\mathbf{a})^2+m(\mathbf{a} \cdot \mathbf{b})+n(\mathbf{a} \cdot(\mathbf{a} \times \mathbf{b}) \\ & \mathbf{a} \cdot \mathbf{c}=l \end{aligned} $

Similarly, $\mathbf{b} \cdot \mathbf{c}=m$

$ \begin{aligned} & \begin{array}{l} \therefore l-m=1 \Rightarrow l^2+m^2-2 l m=1 \\ \text { Now, }(\mathbf{c}-l \mathbf{a}-m \mathbf{b})^2=(n(\mathbf{a} \times \mathbf{b}))^2 \\ \Rightarrow \mathbf{c}^2+l^2 \mathbf{a}^2+m^2 \mathbf{b}-2 m \mathbf{b} \cdot \mathbf{c}+2 l m \mathbf{a} \cdot \mathbf{b}-2 l \mathbf{a} \cdot \mathbf{c}=n^2|\mathbf{a}|^2|\mathbf{b}|^2 \\ \Rightarrow 1+l^2+m^2-2 m^2-2 l^2=n^2 \\ \Rightarrow 1-l^2-m^2=n^2 \\ \Rightarrow n^2=-2 l m \quad\left(\because-2 l m=1-l^2-m^2\right) \end{array} \end{aligned} $

Let $\mathbf{n}$ is the normal to the face $O A B$

$ \because \quad \mathbf{n}=\mathbf{O A} \times \mathbf{O B} $

$ \begin{aligned} & \mathbf{n}=\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & 2 & 1 \\ 2 & 1 & 3 \end{array}\right| \\ & \mathbf{n}=\hat{\mathbf{i}}(5)-\hat{\mathbf{j}}(\mathbf{l})+\hat{\mathbf{k}}(-3) \\ & \mathbf{n}=5 \hat{\mathbf{i}}-\hat{\mathbf{j}}-3 \hat{\mathbf{k}} \end{aligned} $

Direction ratio of edge $\mathbf{B C}=-3,0,-1$

Now, angle between $\mathbf{n}$ and edge $\mathbf{B C}$ is

$ \begin{aligned} && \sin \theta & =\left|\frac{-15-0+3}{\sqrt{35} \sqrt{10}}\right|=\frac{-12}{\sqrt{350}} \times \frac{\sqrt{2}}{\sqrt{2}} \\ \therefore & & \sin \theta & =\frac{6 \sqrt{2}}{5 \sqrt{7}} \\ & \Rightarrow & \theta & =\sin ^{-1}\left(\frac{6 \sqrt{2}}{5 \sqrt{7}}\right) \end{aligned} $

$ \begin{aligned} \mathbf{A B} & =-3 \hat{\mathbf{i}}-3 \hat{\mathbf{k}} \\ \mathbf{B C} & =4 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-4 \hat{\mathbf{k}} \\ \mathbf{C A} & =-\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+7 \hat{\mathbf{k}} \end{aligned} $

Given $\alpha, \beta$ and $\gamma$ are angles between the sides

$ \begin{aligned} \because \cos \alpha= & \frac{\mathbf{A B} \cdot \mathbf{B C}}{|\mathbf{A B}||\mathbf{B C}|}=0 \\ \cos \beta= & \frac{\mathbf{B C} \cdot \mathbf{C A}}{|\mathbf{B C}||\mathbf{C A}|}=\frac{-36}{6 \sqrt{54}}=-\frac{2}{\sqrt{6}} \\ \cos \gamma= & \frac{\mathbf{C A} \cdot \mathbf{A B}}{|\mathbf{C A}||\mathbf{A B}|}=\frac{-18}{6 \sqrt{54}}=\frac{2}{\sqrt{12}} \\ \sin ^2 \alpha+ & \sin ^2 \beta+\sin ^2 \gamma=\sqrt{\left(1-\cos ^2 \alpha\right)} \\ & \quad+\left(\sqrt{1-\cos ^2 \beta}\right)+\left(\sqrt{1-\cos ^2 \gamma}\right) \\ = & 1+\frac{2}{6}+\frac{8}{12}=1+1=2 \end{aligned} $

We have, $A(2 \hat{\mathbf{i}}-\hat{\mathbf{j}}-\hat{\mathbf{k}}), B(5 \hat{\mathbf{i}}+\hat{\mathbf{j}}-2 \hat{\mathbf{k}})$ and $C(-13 \hat{\mathbf{i}}-11 \hat{\mathbf{j}}+4 \hat{\mathbf{k}})$,

$ \begin{aligned} \mathbf{A B} & =\lambda \mathbf{B C} \\ 3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}-\hat{\mathbf{k}} & =\lambda(-18 \hat{\mathbf{i}}-12 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}) \\ \because-18 \lambda & =3 \\ \lambda & =-\frac{3}{18} \end{aligned} $

Given, $\mathbf{A C}=\mu \mathbf{C B}$

$ \begin{aligned} -15 \hat{\mathbf{i}}-10 \hat{\mathbf{j}}+5 \hat{\mathbf{k}} & =\mu(18 \hat{\mathbf{i}}+12 \hat{\mathbf{j}}-6 \hat{\mathbf{k}}) & \\ \mu & =-\frac{15}{18} & \\ \therefore \,\,& \lambda+\mu =-\frac{3}{18}-\frac{15}{18}=-\frac{18}{18}=-1 \end{aligned} $

$ \begin{aligned} &\text { }\\ &\begin{aligned} & A C=O C-O A \\ & O C=O A+A C \\ & O C=a+3 A B \\ & O C=a+3(b-a) \\ & O C=3 b-2 a \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { }\\ &\begin{aligned}Similarly,\,\, \mathbf{B D} & =\mathbf{O D}-\mathbf{O B} \\ \mathbf{O D} & =\mathbf{O B}+\mathbf{B D} \\ & =\mathbf{b}+2 \mathbf{B A} \\ & =\mathbf{b}+2(\mathbf{a}-\mathbf{b}) \\ \mathbf{O D} & =2 \mathbf{a}-\mathbf{b} \end{aligned}\\ &\begin{aligned} C D & =O D-O C \\ & =(2 a-b)-(3 b-2 a) \\ C D & =4 a-4 b \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { Given, }\\ &\begin{aligned} & |\mathbf{a}-\mathbf{b}|^2+|\mathbf{b}-\mathbf{c}|^2+|\mathbf{c}-\mathbf{a}|^2=15 \\ \Rightarrow \quad & |\mathbf{a}|^2+|\mathbf{b}|^2-2 \mathbf{a} \cdot \mathbf{b}+|\mathbf{b}|^2+|\mathbf{c}|^2 \\ & -2 \mathbf{b} \cdot \mathbf{c}+|\mathbf{c}|^2+|\mathbf{a}|^2-2 \mathbf{c} \cdot \mathbf{a}=15 \\ \Rightarrow \quad & 6-2(\mathbf{a} \cdot \mathbf{b}+\mathbf{b} \cdot \mathbf{c}+\mathbf{c} \cdot \mathbf{a})=15 \\ \Rightarrow \quad & \mathbf{a} \cdot \mathbf{b}+\mathbf{b} \cdot \mathbf{c}+\mathbf{c} \cdot \mathbf{a}=-\frac{9}{2} \\ \quad & |\mathbf{a}-\mathbf{b}-\mathbf{c}|^2-4(\mathbf{b} \cdot \mathbf{c}) \\ \Rightarrow \quad & |\mathbf{a}|^2+|\mathbf{b}|^2+|\mathbf{c}|^2 \\ & +2(-\mathbf{a} \cdot \mathbf{b}+\mathbf{b} \cdot \mathbf{c}-\mathbf{c} \cdot \mathbf{a})-4(\mathbf{b} \cdot \mathbf{c}) \\ & =3-2(\mathbf{a} \cdot \mathbf{b}+\mathbf{b} \cdot \mathbf{c}+\mathbf{c} \cdot \mathbf{a}) \\ & =3-2\left(-\frac{9}{2}\right) \\ & =3+9=12 \end{aligned} \end{aligned} $

$\mathbf{a} \times \mathbf{b}=\left|\begin{array}{ccc}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & p & -3 \\ p & -3 & 1\end{array}\right|$

$ \begin{aligned} \hat{\mathbf{i}}(p-9)-\hat{\mathbf{j}}(1+3 p)+k\left(-3-p^2\right) & \\ |\mathbf{a} \times \mathbf{b}|^2 & =(p-9)^2+(3 p+1)^2+\left(p^2+3\right)^2 \\ \mathbf{a} \times \mathbf{c} & =\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & p & -3 \\ -3 & 1 & 2 \end{array}\right| \\ & =\hat{\mathbf{i}}(2 p+3)-\hat{\mathbf{j}}(2-9)+\hat{\mathbf{k}}(1+3 p) \\ |\mathbf{a} \times \mathbf{c}|^2 & =(2 p+3)^2+49+(3 p+1)^2 \end{aligned} $

Given, $|\mathbf{a} \times \mathbf{b}|=|\mathbf{a} \times \mathbf{c}|$

$ \begin{array}{r} \because \quad(p-9)^2+(3 p+1)^2+\left(p^2+3\right)^2=(2 p+3)^2 +49+(3 p+1)^2 \\ \Rightarrow \quad p^2+81-18 p+\left(p^2+3+2 p+3\right) \left(p^2+3-2 p-3\right)=49 \\ \Rightarrow \quad p^2+81-18 p+\left(p^2+2 p+6\right)\left(p^2-2 p\right) =49 \\ \Rightarrow p^2+81-18 p+\left(p^2+2 p+6\right)\left(p^2-2 p\right) =49 \end{array} $

$ \begin{aligned} &\Rightarrow p^4+3 p^2-30 p+32=0\\ &\text { Now put } p=2.16+12-.60+32=0\\ &\because \quad p=2 \end{aligned} $

$(a \times b) \times(c \times d)$

$ \begin{aligned} & =[\mathbf{a} \mathbf{b} \mathbf{d}] \mathbf{c}-[\mathbf{a} \mathbf{b} \mathbf{c}] \mathbf{d} \\ {[\mathbf{a} \mathbf{b} \mathbf{d}] \mathbf{c} } & =8(3 \hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}}) \\ & =24 \hat{\mathbf{i}}-8 \hat{\mathbf{j}}+16 \hat{\mathbf{k}} \\ {[\mathbf{a} \mathbf{b} \mathbf{c}] \mathbf{d} } & =-7(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}) \\ & =-7 \hat{\mathbf{i}}-7 \hat{\mathbf{j}}-7 \hat{\mathbf{k}} \end{aligned} $

$[\mathbf{a} \mathbf{b} \mathbf{d}] \mathbf{c}-[\mathbf{a} \mathbf{b} \mathbf{c}] \mathbf{d}=31 \hat{\mathbf{i}}-\hat{\mathbf{j}}+23 \hat{\mathbf{k}}$

$\begin{aligned} & \because \quad \forall \text { two vectors } \vec{c} \text { & } \vec{d} \\ & |\vec{c} \times \vec{d}|^2=|\vec{c}|^2|\vec{d}|^2-(\vec{c} \cdot \vec{d})^2 \\ & \text { replacing } \vec{c}=\vec{a} ~\& ~\vec{d}=\vec{r} \\ & \Rightarrow|\vec{a} \times \vec{r}|=|\vec{a}|^2|\vec{r}|^2-(\vec{a} \cdot \vec{r})^2 \\ & \Rightarrow|\vec{a} \times \vec{b}|=|\vec{a}|^2|\vec{r}|^2 \quad(\because \vec{a} \times \vec{r}=\vec{a} \times \vec{b} \text { and } \vec{a} \cdot \vec{r}=0) \end{aligned}$

$\begin{aligned} & \Rightarrow 35=14|\vec{r}|^2 \\ & \Rightarrow|\vec{r}|=\sqrt{\frac{35}{14}}=\sqrt{\frac{5}{2}} \neq \sqrt{10} \end{aligned}$

$\therefore$ Statement I is incorrect

Statement II is correct

$\text { (i.e., } \cos 2 A+\cos 2 B+\cos 2 C \geq-\frac{3}{2} \text { ) }$

Proof: $\because(\overrightarrow{O A}+\overrightarrow{O B}+\overrightarrow{O C}) \geq 0\quad \text{..... (1)}$

and $|\overrightarrow{O A}|^2=|\overrightarrow{O B}|^2=|\overrightarrow{O C}|^2=R^2\quad \text{..... (2)}$

Now, using (1), we get

$|\overrightarrow{O A}|^2+|\overrightarrow{O B}|^2+|\overrightarrow{O C}|^2 +2(\overrightarrow{O A} \cdot \overrightarrow{O B}+\overrightarrow{O B} \cdot \overrightarrow{O C}+\overrightarrow{O C} \cdot \overrightarrow{O A}) \geq 0$

$ \begin{aligned} & \Rightarrow 3 R^2+2 R^2(\cos 2 A+\cos 2 B+\cos 2 C) \geq 0 \\ & \Rightarrow \cos 2 A+\cos 2 B+\cos 2 C \geq-\frac{3}{2} \end{aligned} $

Area of parallelogram whose diagonals are $\vec{a}+\vec{b}$ and $\vec{b}+\vec{c}$ is

$\begin{aligned} & =\frac{1}{2}|(\vec{a}+\vec{b}) \times(\vec{b}+\vec{c})| \\ & =\frac{1}{2}|\vec{a} \times \vec{b}+\vec{a} \times \vec{c}+\vec{b} \times \vec{c}| \\ & =\frac{1}{2}|-2 \beta \hat{i}-2 \hat{j}+(\alpha+\beta) \hat{k}| \end{aligned}$

$=\frac{1}{2} \sqrt{4 \beta^2+4+(\alpha+\beta)^2}$

Which is given $\frac{\sqrt{21}}{2}$

$\begin{array}{ll} \therefore & 4 \beta^2+4+(\alpha+\beta)^2=21 \\ \Rightarrow & (\alpha+\beta)^2+4 \beta^2=17 \\ \Rightarrow & \alpha^2+5 \beta^2+2 \alpha \beta=17 \\ \Rightarrow & \alpha^2+5 \beta^2=29 \\ \therefore & (\alpha, \beta) \in\{(3,2),(-3,-2),(-3,2),(3,-2)\} \\ \because & \alpha \beta=-6 \\ \therefore & (\alpha, \beta) \in\{(-3,2),(3,-2)\} \\ \therefore & \alpha_1^2+\beta_1^2-\alpha_2 \beta_2 \\ = & 9+4-(-6)=19 \end{array}$

To solve this, let's start with the given vector equation:

$\vec{c} = \vec{a} - \vec{b}$

Given vectors are:

$\overrightarrow{\mathrm{a}} = \alpha \hat{i} + 4 \hat{j} + 2 \hat{k}$

$\overrightarrow{\mathrm{b}} = 5 \hat{i} + 3 \hat{j} + 4 \hat{k}$

Then, the vector $\overrightarrow{\mathrm{c}}$ is:

$\overrightarrow{\mathrm{c}} = (\alpha \hat{i} + 4 \hat{j} + 2 \hat{k}) - (5 \hat{i} + 3 \hat{j} + 4 \hat{k})$

$\overrightarrow{\mathrm{c}} = (\alpha - 5) \hat{i} + (4 - 3) \hat{j} + (2 - 4) \hat{k}$

$\overrightarrow{\mathrm{c}} = (\alpha - 5) \hat{i} + 1 \hat{j} - 2 \hat{k}$

The area of the triangle formed by vectors $\vec{a}$ and $\vec{b}$ is given by the magnitude of the cross product of $\vec{a}$ and $\vec{b}$, divided by 2:

$\text{Area} = \frac{1}{2} |\vec{a} \times \vec{b}| = 5 \sqrt{6}$

This implies:

$|\vec{a} \times \vec{b}| = 10 \sqrt{6}$

Let's find $\vec{a} \times \vec{b}$:

$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \alpha & 4 & 2 \\ 5 & 3 & 4 \end{vmatrix} = \hat{i}(4 \cdot 4 - 2 \cdot 3) - \hat{j}(\alpha \cdot 4 - 2 \cdot 5) + \hat{k}(\alpha \cdot 3 - 4 \cdot 5)$

$\vec{a} \times \vec{b} = \hat{i}(16 - 6) - \hat{j}(4\alpha - 10) + \hat{k}(3\alpha - 20)$

$\vec{a} \times \vec{b} = \hat{i}(10) - \hat{j}(4\alpha - 10) + \hat{k}(3\alpha - 20)$

Magnitude of $\vec{a} \times \vec{b}$:

$|\vec{a} \times \vec{b}| = \sqrt{10^2 + (4\alpha - 10)^2 + (3\alpha - 20)^2}$

We know:

$\sqrt{10^2 + (4\alpha - 10)^2 + (3\alpha - 20)^2} = 10 \sqrt{6}$

Squaring both sides:

$100 + (4\alpha - 10)^2 + (3\alpha - 20)^2 = 600$

$100 + 16\alpha^2 - 80\alpha + 100 + 9\alpha^2 - 120\alpha + 400 = 600$

$25\alpha^2 - 200\alpha + 600 = 600$

$25\alpha^2 - 200\alpha = 0$

$\alpha^2 - 8\alpha = 0$

$\alpha(\alpha - 8) = 0$

Since $\alpha$ is a positive real number:

$\alpha = 8$

Then, the vector $\vec{c}$ is:

$\vec{c} = (8 - 5)\hat{i} + 1\hat{j} - 2\hat{k}$

$\vec{c} = 3\hat{i} + 1\hat{j} - 2\hat{k}$

Magnitude squared of $\vec{c}$ is:

$|\vec{c}|^2 = 3^2 + 1^2 + (-2)^2$

$|\vec{c}|^2 = 9 + 1 + 4$

$|\vec{c}|^2 = 14$

Therefore, the correct option is:

Option A: 14

$\begin{aligned} & 6|\vec{a} \times \vec{b}|=15 \\ & \Rightarrow|\vec{a} \times \vec{b}|=\frac{5}{2} \end{aligned}$

Area of quadrilateral $O A B C$

$=$ area of $\triangle O A C+$ area of $\triangle A B C$

$\begin{aligned} & =\frac{15}{2}+\frac{1}{2}|(\overrightarrow{A B} \times \overrightarrow{B C})| \\ & =\frac{15}{2}+\frac{1}{2}|(4 \vec{a}+5 \vec{b}) \times(6 \vec{a}+2 \vec{b})| \\ & =\frac{15}{2}+\frac{1}{2}|(22 \vec{a} \times \vec{b})| \\ & =\frac{15}{2}+11 \times \frac{5}{2} \\ & =\frac{70}{2}=35 \end{aligned}$

$\begin{aligned} & \left.\begin{array}{l} \vec{a}=4 \hat{i}-\hat{j}+\hat{k} \\ \vec{b}=11 \hat{i}-\hat{j}+\hat{k} \end{array}\right] \\ & \vec{a} \cdot \vec{b}=44+1+1=46 \\ & |\vec{a}|^2=18,\left|\vec{b}^2\right|=123 \\ & (\vec{a}+\vec{b}) \times \vec{c}=\vec{c} \times(-2 \vec{a}+3 \vec{b}) \\ & (2 \vec{a}+3 \vec{b}) \cdot \vec{c}=1670 \\ & (\vec{a}+\vec{b}) \times c=(2 \vec{a}-3 \vec{b}) \times \vec{c} \\ & (-\vec{a}+4 \vec{b}) \times \vec{c}=0 \\ & \vec{c}=\lambda(4 \vec{b}-\vec{a}) \\ & (2 \vec{a}+3 \vec{b}) \cdot \lambda(4 \vec{b}-\vec{a})=1670 \\ & \lambda\left(5 \vec{a} \cdot \vec{b}-2|\vec{a}|^2+12|\vec{b}|^2\right)=1670 \\ & \lambda=\frac{1670}{5 \times 46-2 \times 18+12 \times 123} \\ & \lambda=1 \\ & \vec{c}=4 \vec{b}-\vec{a} \\ & =4(11 \hat{i}-\hat{j}+\hat{k})-(4 \hat{i}-\hat{j}+\hat{k}) \\ & =40 \hat{i}-3 \hat{j}+3 \hat{k} \\ & \left|\vec{c}^2\right|=1600+9+9 \\ & =1618 \\ \end{aligned}$

$\begin{aligned} & \vec{a}=\hat{i}+2 \hat{j}+3 \hat{k} \\ & \vec{b}=2 \hat{i}+3 \hat{j}-5 \hat{k} \\ & \vec{c}=3 \hat{i}-\hat{j}+\lambda \hat{k} \\ & \vec{b}+\vec{c}=5 \hat{i}+2 \hat{j}+(\lambda-5) \hat{k} \end{aligned}$

$\vec{r}$ is a unit vector along $\vec{b}+\vec{c}$

$\therefore \quad \vec{r}=\frac{5 \hat{i}+2 \hat{j}+(\lambda-5) \hat{k}}{\sqrt{25+4+(\lambda-5)^2}}$

Now, $\vec{r} \cdot \vec{a}=3$

$\frac{1}{\sqrt{29+(\lambda-5)^2}}[5+4+3(\lambda-5)]=3$

Squaring both sides

$\begin{aligned} & \Rightarrow \quad \frac{1}{29+(\lambda-5)^2}\left[9+3(\lambda-5)^2\right]=9 \\ & \Rightarrow \quad[3+(\lambda-5)]^2=29+(\lambda-5)^2 \\ & \Rightarrow \quad 9+(\lambda-5)^2+6(\lambda-5)=29+(\lambda-5)^2 \\ & \Rightarrow \quad 9+6(\lambda-5)=29 \\ & \Rightarrow \quad \lambda=\frac{20}{6}+5=\frac{25}{3} \\ & \therefore \quad 3 \lambda=3 \times \frac{25}{3}=25 \end{aligned}$

Given $\vec{a}=\alpha t \hat{i}+6 \hat{j}-3 \hat{k}$

and $\vec{b}=t \hat{i}-2 \hat{j}-2 \alpha t \hat{k}$

angle between $\vec{a}$ and $\vec{b}$ is given by

$\cos \theta=\frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}$

We have, $\cos \theta < 0(\because$ angle between $\vec{a}$ and $\vec{b}$ is obtuse)

$\begin{aligned} & \Rightarrow \quad \vec{a} \cdot \vec{b}<0 \\ & \Rightarrow \alpha t^2-12+6 \alpha t<0 \forall t \in \mathbb{R} \end{aligned}$

If $\alpha=0$, then $-12<0$ (condition holds)

If $\alpha \neq 0 \Rightarrow \alpha<0\quad \text{.... (i)}$

And maximum value of $\alpha t^2+6 \alpha t-12<0$

$\begin{aligned} & \Rightarrow \frac{-D}{4 a}<0 \text { (where } D \text { is discriminant and } a=\alpha) \\ & \Rightarrow \frac{36 \alpha^2+48 \alpha}{4 \alpha}>0 \\ & \Rightarrow \alpha>\frac{-4}{3} \\ & \therefore \alpha \in\left(\frac{-4}{3}, 0\right] \end{aligned}$

$\begin{aligned} & \vec{a}=2 \hat{i}+\hat{j}-\hat{k} \\ & \vec{b}=((\vec{a} \times(\hat{i}+\hat{j})) \times \hat{i}) \times \hat{i} \\ & \vec{a} \times(\hat{i}+\hat{j})=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -1 \\ 1 & 1 & 0 \end{array}\right| \\ &=\hat{i}(1)-\hat{j}(1)+\hat{k}(2-1) \\ &=\hat{i}-\hat{j}+\hat{k} \end{aligned}$

$\begin{aligned} & (\vec{a} \times(\hat{i}+\hat{j})) \times \hat{i}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 1 \\ 1 & 0 & 0 \end{array}\right| \\ & \quad=\hat{i}(0)-\hat{j}(-1)+\hat{k}(1) \\ & \quad=\hat{j}+\hat{k} \end{aligned}$

$\left( {(\overrightarrow a \times (\widehat i + \widehat j) \times \widehat i} \right) \times \widehat i = \left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 0 & 1 & 1 \cr 1 & 0 & 0 \cr } } \right|$

$\begin{aligned} & =\hat{i}(0)-\hat{j}(-1)+\hat{k}(-1) \\ \vec{b}= & \hat{j}-\hat{k} \end{aligned}$

Projection of $\vec{a}$ on $\vec{b}=\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$

$=\frac{2}{\sqrt{2}}=\sqrt{2}$

Square of projection $=2$

$\begin{aligned} & |(\vec{a} \times \vec{b}) \times \vec{c}|=|\vec{a} \times \vec{b}||\vec{c}| \sin 60^{\circ} \\ & \left|\begin{array}{ccc} i & j & k \\ 6 & 1 & -1 \\ 1 & 1 & 0 \end{array}\right|=i(1)-j(1)+k(5) \\ & =i-j+5 k \\ & |\vec{a} \times \vec{b}|=\sqrt{1+1+25}=\sqrt{27} \\ & |\vec{c}-\vec{a}|=2 \sqrt{2} \\ & c^2+a^2-2 a c=8 \\ & c^2-12 c+30=0 \\ & c=\frac{12+\sqrt{24}}{2}=6+\sqrt{6} \\ & \Rightarrow|(\vec{a} \times \vec{b}) \times \vec{c}|=\sqrt{27} \times(6+\sqrt{6}) \times \frac{\sqrt{3}}{2} \\ & =\frac{a}{2}(6+\sqrt{6}) \\ \end{aligned}$

$\begin{gathered} (\vec{c}+\hat{i}) \times(\vec{a}+\vec{b}+\hat{i}+\vec{a})=0 \\ \Rightarrow \quad \vec{c}+\hat{i}=\lambda(\vec{a}+\vec{b}+\hat{i}+\vec{a}) \\ =\lambda(2 \vec{a}+\vec{b}+\hat{i}) \\ \quad=\lambda(7 \hat{i}+8 \hat{j}) \\ \Rightarrow \quad \vec{c}=(7 \lambda-1) \hat{i}+8 \lambda \hat{j} \\ \quad \vec{c} \cdot \vec{a}=-29 \\ \Rightarrow \quad 14 \lambda-2+40 \lambda=-29 \\ \Rightarrow \quad 54 \lambda=-27 \\ \Rightarrow \quad \lambda=-\frac{1}{2} \end{gathered}$

$\begin{aligned} & \therefore \quad \vec{c}=\left(\frac{-7}{2}-1\right) \hat{i}-4 \hat{j}=\frac{-9}{2} \hat{i}-4 \hat{j} \\ & \vec{c} \cdot(-2 \hat{i}+\hat{j}+\hat{k})=9-4=5 \end{aligned}$

$\begin{aligned} & \vec{a}=\vec{b} \times \vec{c} \\ & |\vec{a}|=2,|\vec{b}|=3 \end{aligned}$

$\vec{a} \cdot \vec{b}=0$ and $\vec{a} \cdot \vec{c}=0$

$\begin{aligned} & |\vec{c}-\vec{a}|^2=|\vec{c}|^2+|\vec{a}|^2-2 \vec{c} \cdot \vec{a} \\ & =4+|\vec{c}|^2 \end{aligned}$

$\begin{aligned} & |\vec{a}|=|\vec{b} \times \vec{c}|=|\vec{b}| \sin \alpha|\vec{c}| \\ & \Rightarrow \sin \alpha|\vec{c}|=\frac{2}{3} \\ & \Rightarrow \sin ^2 \alpha=\frac{4}{9|\vec{c}|^2} \\ & \Rightarrow|\vec{c}|^2=\frac{4}{9 \sin ^2 \alpha} \\ & \Rightarrow|\vec{c}-\vec{a}|^2=4+\frac{4}{9 \sin ^2 \alpha} \end{aligned}$

For $|\vec{c}-\vec{a}|^2$ to be minimum for $\alpha \in\left[0, \frac{\pi}{3}\right]$

$\begin{gathered} \sin \alpha=\frac{\sqrt{3}}{2} \\ 27|\vec{c}-\vec{a}|^2=27\left[4+\frac{4.4}{9.3}\right] \\ =27\left[\frac{124}{27}\right]=124 \end{gathered}$

Area of quadrilateral $A B C D=$ area of $\triangle A B C+$ area of $\triangle A D C$

In $\triangle A B C$

$\begin{aligned} & \overrightarrow{A B}=4,8,-8 \\ & \overrightarrow{B C}=-2,-3,-4 \\ & \overrightarrow{A B} \times \overrightarrow{B C}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 4 & 8 & -8 \\ -2 & -3 & -4 \end{array}\right| \\ & =-56 \hat{i}+32 \hat{j}+4 \hat{k} \end{aligned}$

Area of $\triangle A B C=\frac{1}{2}|\overrightarrow{A B} \times \overrightarrow{B C}|$

$\begin{aligned} & =\frac{1}{2} \sqrt{56^2+32^2+4^2} \\ & =\frac{1}{2} \sqrt{4176}=\frac{12 \sqrt{29}}{2}=6 \sqrt{29} \end{aligned}$

In $\triangle A D C=\overrightarrow{A D}=-2,-3,-4$

$\overrightarrow{D C}=4,8,-8$

$\overrightarrow{A D} \times \overrightarrow{D C}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ -2 & -3 & -4 \\ 4 & 8 & -8\end{array}\right|$

$=56 \hat{i}-32 \hat{j}-4 k$

$\text { Area of } \frac{1}{2}|\overrightarrow{A D} \times \overrightarrow{D C}|$

$\begin{aligned} & =\frac{1}{2} \sqrt{4176} \\ & =6 \sqrt{29} \end{aligned}$

Area of $A B C D=6 \sqrt{29}+6 \sqrt{29}$

$=12 \sqrt{29}$

We are given the vectors

$ \vec{a} = \hat{i} + \lambda\,\hat{j} - 3\,\hat{k} \quad \text{and} \quad \vec{b} = 3\,\hat{i} - \hat{j} + 2\,\hat{k}, $

with $\lambda > 0$, and we are told that the vectors $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b}$ are perpendicular. Our goal is to find $(14 \cos\theta)^2$, where $\theta$ is the angle between $\vec{a}$ and $\vec{b}$.

Step 1. Use the Perpendicularity Condition

Two vectors are perpendicular if their dot product is zero. So, we require:

$ (\vec{a}+\vec{b})\cdot (\vec{a}-\vec{b})=0. $

First, compute $\vec{a}+\vec{b}$ and $\vec{a}-\vec{b}$:

$ \vec{a}+\vec{b} = (\hat{i}+3\,\hat{i})+(\lambda\,\hat{j}-\hat{j})+(-3\,\hat{k}+2\,\hat{k}) = 4\,\hat{i}+(\lambda-1)\,\hat{j}-\hat{k}. $

$ \vec{a}-\vec{b} = (\hat{i}-3\,\hat{i})+(\lambda\,\hat{j}+ \hat{j})+(-3\,\hat{k}-2\,\hat{k}) = -2\,\hat{i}+(\lambda+1)\,\hat{j}-5\,\hat{k}. $

Now, take their dot product:

$ (4\,\hat{i}+(\lambda-1)\,\hat{j}-\hat{k}) \cdot (-2\,\hat{i}+(\lambda+1)\,\hat{j}-5\,\hat{k}) $

$ = 4(-2) + (\lambda-1)(\lambda+1) + (-1)(-5) = -8 + (\lambda^2 - 1) + 5. $

$ = \lambda^2 - 4. $

Setting the dot product to zero:

$ \lambda^2 - 4 = 0 \quad \Longrightarrow \quad \lambda^2 = 4. $

Since $\lambda > 0$, we have:

$ \lambda = 2. $

Step 2. Find the Angle Between $\vec{a}$ and $\vec{b}$

Now that we know $\lambda = 2$, the vectors become:

$ \vec{a} = \hat{i} + 2\,\hat{j} - 3\,\hat{k}, \quad \vec{b} = 3\,\hat{i} - \hat{j} + 2\,\hat{k}. $

The cosine of the angle $\theta$ between $\vec{a}$ and $\vec{b}$ is given by:

$ \cos\theta = \frac{\vec{a}\cdot\vec{b}}{\|\vec{a}\| \|\vec{b}\|}. $

Calculate $\vec{a} \cdot \vec{b}$:

$ \vec{a}\cdot\vec{b} = (1)(3) + (2)(-1) + (-3)(2) = 3 - 2 - 6 = -5. $

Calculate the magnitudes:

$ \|\vec{a}\| = \sqrt{1^2 + 2^2 + (-3)^2} = \sqrt{1 + 4 + 9} = \sqrt{14}. $

$ \|\vec{b}\| = \sqrt{3^2 + (-1)^2 + 2^2} = \sqrt{9 + 1 + 4} = \sqrt{14}. $

Thus,

$ \cos\theta = \frac{-5}{\sqrt{14}\cdot\sqrt{14}} = \frac{-5}{14}. $

Step 3. Compute $(14\cos\theta)^2$

$ (14\cos\theta)^2 = \left(14\cdot\frac{-5}{14}\right)^2 = (-5)^2 = 25. $

Final Answer

The value of $(14\cos\theta)^2$ is:

$ \boxed{25}. $

$\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{b}=2 \hat{i}+4 \hat{j}-5 \hat{k}, \vec{c}=x \hat{i}+2 \hat{j}+3 \hat{k}, x \in R$

$\text { also, } \vec{b}+\vec{c}=(x+2) \hat{i}+6 \hat{j}-2 \hat{k}$

$\vec{d} \text { is the unit vector in the direction of } \vec{b}+\vec{c}$

$\begin{aligned} & |\vec{b}+\vec{c}|=\sqrt{(x+2)^2+6^2+2^2} \\ & =\sqrt{40+(x+2)^2} \\ & \vec{d}=\frac{x+2}{\sqrt{40+(x+2)^2}} \hat{i}+\frac{6}{\sqrt{40+(x+2)^2}} \hat{j} -\frac{2}{\sqrt{40+(x+2)^2}} \hat{k} \\ \end{aligned}$

$\begin{aligned} & \vec{a} \cdot \vec{d}=1 \\ & \frac{x+2+6-2}{\sqrt{40+(x+2)^2}}=1 \\ & x+6=\sqrt{40+(x+2)^2} \end{aligned}$

$\begin{aligned} &(x+6)^2=40+(x+2)^2 \\ & x^2+36+ 12 x=40+x^2+4+4 x \\ & 8 x=8 \\ & \Rightarrow x=1 \\ &(\vec{a} \times \vec{b}) \cdot \vec{c}=[\vec{a} \vec{b} \vec{c}] \\ &=\left[\begin{array}{ccc} 1 & 1 & 1 \\ 2 & 4 & -5 \\ 1 & 2 & 3 \end{array}\right] \\ &=1(12+10)-1(6+5)+1(4-4) \\ &=22-11=11 \end{aligned}$

$\begin{aligned} & \text { Let } \vec{C}=a \hat{i}+b \hat{j}+c \hat{k} \\ & (a \hat{i}+b \hat{j}+c \hat{k}) \cdot(2 \hat{i}+2 \hat{j}-\hat{k})=1 \times 3 \times \frac{1}{2} \\ & 2 a+2 b-c=\frac{3}{2} \qquad \text{... (1)}\\ & (a \hat{i}+b \hat{j}+c \hat{k}) \cdot(\hat{i}-\hat{k})=1 \times \sqrt{2} \times \frac{1}{\sqrt{2}} \\ & a-c=1 \quad \text{... (2)}\\ & a^2+b^2+c^2=1 \quad \text{... (3)}\\ \end{aligned}$

Solving (1), (2) and (3)

$\begin{aligned} & a+2 b=\frac{1}{2} \\ & a^2+b^2+(a-1)^2=1 \\ & 2 a^2-2 a+b^2=0 \\ & 2 a^2-2 a+\left(\frac{2 a-1}{4}\right)^2=0 \\ & 32 a^2-32 a+4 a^2-4 a+1=0 \\ & 36 a^2-36 a+1=0 \\ & a=\frac{36 \pm \sqrt{(36)^2-4(36)}}{2 \times 36} \\ & =\frac{1}{2} \pm \frac{\sqrt{2}}{3} \\ & b=\frac{1-2 a}{4} \Rightarrow b=\frac{1 \pm \frac{2 \sqrt{2}}{3}-1}{4} \\ & =\mp \frac{1}{3 \sqrt{2}} \\ & C=-\frac{1}{2} \pm \frac{\sqrt{2}}{3} \end{aligned}$

$\begin{aligned} & C+\left(\frac{-1}{2} \hat{i}+\frac{1}{3 \sqrt{2}} \hat{j}-\frac{\sqrt{2}}{3} \hat{k}\right) \\ & =\frac{\sqrt{2}}{3} \hat{i}-\frac{1}{2} \hat{k} \end{aligned}$

$\begin{aligned} & \overrightarrow{\mathrm{p}} \times \overrightarrow{\mathrm{b}}-\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{b}}=\overrightarrow{0} \\ & (\overrightarrow{\mathrm{p}}-\overrightarrow{\mathrm{c}}) \times \overrightarrow{\mathrm{b}}=\overrightarrow{0} \\ & \overrightarrow{\mathrm{p}}-\overrightarrow{\mathrm{c}}=\lambda \overrightarrow{\mathrm{b}} \Rightarrow \overrightarrow{\mathrm{p}}=\overrightarrow{\mathrm{c}}+\lambda \overrightarrow{\mathrm{b}} \end{aligned}$

Now, $\overrightarrow{\mathrm{p}} \cdot \overrightarrow{\mathrm{a}}=0$ (given)

$\begin{aligned} & \text { So, } \overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{a}}+\lambda \overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}=0 \\ & (3-3-8)+\lambda(12+1-14)=0 \\ & \lambda=-8 \\ & \overrightarrow{\mathrm{p}}=\overrightarrow{\mathrm{c}}-8 \overrightarrow{\mathrm{b}} \\ & \overrightarrow{\mathrm{p}}=-31 \hat{\mathrm{i}}-11 \hat{\mathrm{j}}-52 \hat{\mathrm{k}} \\ & \text { So, } \overrightarrow{\mathrm{p}} \cdot(\hat{\mathrm{i}}-\hat{\mathrm{j}}-\hat{\mathrm{k}}) \\ & =-31+11+52 \\ & =32 \end{aligned}$

A vector in the direction of the required line can be obtained by cross product of

$\begin{aligned} & \left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 5 \\ -1 & 3 & 2 \end{array}\right| \\\\ & =-9 \hat{i}-9 \hat{j}+9 \hat{k} \end{aligned}$

Required line

$\begin{aligned} & \overrightarrow{\mathrm{r}}=(5 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})+\lambda^{\prime}(-9 \hat{\mathrm{i}}-9 \hat{\mathrm{j}}+9 \hat{\mathrm{k}}) \\\\ & \overrightarrow{\mathrm{r}}=(5 \hat{\mathrm{i}}-4 \hat{\mathrm{j}}+3 \hat{\mathrm{k}})+\lambda(\hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}}) \end{aligned}$

Now distance of $(0,2,-2)$

$\begin{aligned} & \text { P.V. of } \mathrm{P} \equiv(5+\lambda) \hat{i}+(\lambda-4) \hat{j}+(3-\lambda) \hat{k} \\\\ & \overrightarrow{\mathrm{AP}}=(5+\lambda) \hat{i}+(\lambda-6) \hat{j}+(5-\lambda) \hat{k} \\\\ & \overrightarrow{\mathrm{AP}} \cdot(\hat{\mathrm{i}}+\hat{\mathrm{j}}-\hat{\mathrm{k}})=0 \\\\ & 5+\lambda+\lambda-6-5+\lambda=0 \\\\ & \lambda=2 \\\\ & |\overrightarrow{\mathrm{AP}}|=\sqrt{49+16+9} \\\\ & |\overrightarrow{\mathrm{AP}}|=\sqrt{74} \end{aligned}$

$\begin{aligned} & |\overrightarrow{\mathrm{b}}|^2=6 ;|\overrightarrow{\mathrm{a}}||\overrightarrow{\mathrm{b}}| \cos \theta=3 \sqrt{2} \\ & |\overrightarrow{\mathrm{a}}|^2|\overrightarrow{\mathrm{b}}|^2 \cos ^2 \theta=18 \\ & |\overrightarrow{\mathrm{a}}|^2=6 \end{aligned}$

Also $1+\alpha^2+\beta^2=6$

$\alpha^2+\beta^2=5$

to find

$\begin{aligned} & \left(\alpha^2+\beta^2\right)|\vec{a}|^2|\vec{b}|^2 \sin ^2 \theta \\ & =(5)(6)(6)\left(\frac{1}{2}\right) \\ & =90 \end{aligned}$

To find the value of $|(\vec{b} \times \vec{a})-\vec{b}|^2$, we can use properties of vector operations and magnitudes. Given $|\vec{b}| = 1$ and $|\vec{b} \times \vec{a}| = 2$, let's break down the calculation step by step:

Firstly, we observe that the cross product of two vectors $\vec{b} \times \vec{a}$ is orthogonal (perpendicular) to both $\vec{b}$ and $\vec{a}$. This means that when we take the dot product of $\vec{b} \times \vec{a}$ with either $\vec{b}$ or $\vec{a}$, the result will be zero due to the orthogonal property. Specifically,

$(\vec{b} \times \vec{a}) \cdot \vec{b}= 0 $(because $\vec{b} \times \vec{a}$ is perpendicular to both $\vec{b}$ and $\vec{a}$)

Next, to find the magnitude squared of the vector $(\vec{b} \times \vec{a}) - \vec{b}$, we apply the formula for the magnitude squared of a vector subtraction, which can be expressed as:

$|(\vec{b} \times \vec{a}) - \vec{b}|^2 = |\vec{b} \times \vec{a}|^2 + |-\vec{b}|^2 + 2(\vec{b} \times \vec{a}) \cdot (-\vec{b})$

Since $(\vec{b} \times \vec{a}) \cdot \vec{b} = 0$ and $|-\vec{b}| = |\vec{b}|$, the equation simplifies to:

$= |\vec{b} \times \vec{a}|^2 + |\vec{b}|^2$

Given that $|\vec{b} \times \vec{a}| = 2$ and $|\vec{b}| = 1$, substituting these values gives:

$= 2^2 + 1^2 = 4 + 1 = 5$

Therefore, the value of $|(\vec{b} \times \vec{a})-\vec{b}|^2$ is 5.

Given $|\vec{a}|=1,|\vec{b}|=4, \vec{a} \cdot \vec{b}=2$

$\vec{\mathrm{c}}=2(\vec{\mathrm{a}} \times \vec{\mathrm{b}})-3 \vec{\mathrm{b}}$

Dot product with $\overrightarrow{\mathrm{a}}$ on both sides

$\overrightarrow{\mathrm{c}} . \overrightarrow{\mathrm{a}}=-6$ ..... (1)

Dot product with $\vec{b}$ on both sides

$\begin{aligned} & \overrightarrow{\mathrm{b}} \overrightarrow{\mathrm{c}}=-48 \quad \text{... (2)}\\ & \overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{c}}=4|\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}|^2+9|\overrightarrow{\mathrm{b}}|^2 \\ & |\overrightarrow{\mathrm{c}}|^2=4\left[|\mathrm{a}|^2|\mathrm{~b}|^2-(\mathrm{a} \cdot \overrightarrow{\mathrm{b}})^2\right]+9|\overrightarrow{\mathrm{b}}|^2 \\ & |\overrightarrow{\mathrm{c}}|^2=4\left[(1)(4)^2-(4)\right]+9(16) \\ & |\overrightarrow{\mathrm{c}}|^2=4[12]+144 \\ & |\overrightarrow{\mathrm{c}}|^2=48+144 \\ & |\overrightarrow{\mathrm{c}}|^2=192 \\ & \therefore \cos \theta=\frac{\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}}}{|\overrightarrow{\mathrm{b}}| \overrightarrow{\mathrm{c}} \mid} \\ & \therefore \cos \theta=\frac{-48}{\sqrt{192} \cdot 4} \\ & \therefore \cos \theta=\frac{-48}{8 \sqrt{3} .4} \\ & \therefore \cos \theta=\frac{-3}{2 \sqrt{3}} \\ & \therefore \cos \theta=\frac{-\sqrt{3}}{2} \Rightarrow \theta=\cos ^{-1}\left(\frac{-\sqrt{3}}{2}\right) \end{aligned}$

Unit vector $\hat{\mathrm{u}}=\mathrm{x} \hat{\mathrm{i}}+\mathrm{y} \hat{\mathrm{j}}+\mathrm{z} \hat{\mathrm{k}}$

$\begin{aligned} & \overrightarrow{\mathrm{p}}_1=\frac{1}{\sqrt{2}} \hat{\mathrm{i}}+\frac{1}{\sqrt{2}} \hat{\mathrm{k}}, \overrightarrow{\mathrm{p}}_2=\frac{1}{\sqrt{2}} \hat{\mathrm{j}}+\frac{1}{\sqrt{2}} \hat{\mathrm{k}} \\ & \overrightarrow{\mathrm{p}}_3=\frac{1}{\sqrt{2}} \hat{\mathrm{i}}+\frac{1}{\sqrt{2}} \hat{\mathrm{j}} \end{aligned}$

Now angle between $\hat{\mathrm{u}}$ and $\overrightarrow{\mathrm{p}}_1=\frac{\pi}{2}$

$\hat{\mathrm{u}} \cdot \overrightarrow{\mathrm{p}}_1=0 \Rightarrow \frac{\mathrm{x}}{\sqrt{2}}+\frac{\mathrm{z}}{\sqrt{2}}=0$

$\Rightarrow \mathrm{x}+\mathrm{z}=0$ ...... (i)

Angle between $\hat{\mathrm{u}}$ and $\overrightarrow{\mathrm{p}}_2=\frac{\pi}{3}$

$\hat{\mathrm{u}} \cdot \overrightarrow{\mathrm{p}}_2=|\hat{\mathrm{u}}| \cdot\left|\overrightarrow{\mathrm{p}}_2\right| \cos \frac{\pi}{3}$

$\Rightarrow \frac{y}{\sqrt{2}}+\frac{z}{\sqrt{2}}=\frac{1}{2} \Rightarrow y+z=\frac{1}{\sqrt{2}}$ ...... (ii)

Angle between $\hat{\mathrm{u}}$ and $\overrightarrow{\mathrm{p}}_3=\frac{2 \pi}{3}$

$\hat{\mathrm{u}} \cdot \overrightarrow{\mathrm{p}}_3=|\hat{\mathrm{u}}| \cdot\left|\overrightarrow{\mathrm{p}}_3\right| \cos \frac{2 \pi}{3}$

$\Rightarrow \frac{x}{\sqrt{2}}+\frac{4}{\sqrt{2}}=\frac{-1}{2} \Rightarrow x+y=\frac{-1}{\sqrt{2}}$ ..... (iii)

from equation (i), (ii) and (iii) we get

$x=\frac{-1}{\sqrt{2}} \quad y=0 \quad z=\frac{1}{\sqrt{2}}$

$\begin{aligned} & \text { Thus } \hat{\mathrm{u}}-\overrightarrow{\mathrm{v}}=\frac{-1}{\sqrt{2}} \hat{\mathrm{i}}+\frac{1}{\sqrt{2}} \hat{\mathrm{k}}-\frac{1}{\sqrt{2}} \hat{\mathrm{i}}-\frac{1}{\sqrt{2}} \hat{\mathrm{j}}-\frac{1}{\sqrt{2}} \hat{\mathrm{k}} \\ & \hat{\mathrm{u}}-\overrightarrow{\mathrm{v}}=\frac{-2}{\sqrt{2}} \hat{\mathrm{i}}-\frac{1}{\sqrt{2}} \hat{\mathrm{j}} \\ & \therefore|\hat{\mathrm{u}}-\overrightarrow{\mathrm{v}}|^2=\left(\sqrt{\frac{4}{2}+\frac{1}{2}}\right)^2=\frac{5}{2} \end{aligned}$

Area of parallelogram, $S=|\vec{a} \times \vec{b}|$

Area of quadrilateral $=\operatorname{Area}(\triangle \mathrm{OAB})+\operatorname{Area}(\triangle \mathrm{OBC})$

$\begin{aligned} & =\frac{1}{2}\{|\vec{a} \times(12 \vec{a}+4 \vec{b})|+|\vec{b} \times(12 \vec{a}+4 \vec{b})|\} \\ & =8|(\vec{a} \times \vec{b})| \end{aligned}$

$\text { Ratio }=\frac{8|(\vec{a} \times \vec{b})|}{|(\vec{a} \times \vec{b})|}=8$

$\begin{aligned} & \vec{a}+5 \vec{b}=\lambda \vec{c} \\ & \vec{b}+6 \vec{c}=\mu \vec{a} \end{aligned}$

Eliminating $\vec{a}$

$\begin{aligned} & \lambda \overrightarrow{\mathrm{c}}-5 \overrightarrow{\mathrm{b}}=\frac{6}{\mu} \overrightarrow{\mathrm{c}}+\frac{1}{\mu} \overrightarrow{\mathrm{b}} \\ & \therefore \mu=\frac{-1}{5}, \lambda=-30 \\ & \alpha=5, \beta=30 \end{aligned}$

$\triangle \mathrm{ABC}$ is equilateral

Orthocentre and centroid will be same

$\mathrm{G}\left(\frac{5}{3}, \frac{5}{3}, \frac{5}{3}\right)$

Mid-point of $\mathrm{AB}$ is $\mathrm{D}\left(\frac{3}{2}, 2, \frac{3}{2}\right)$

$\begin{aligned} & \therefore \ell_1=\sqrt{\frac{1}{36}+\frac{1}{9}+\frac{1}{36}} \\ & \ell_1=\sqrt{\frac{1}{6}}=\ell_2=\ell_3 \\ & \therefore \ell_1^2+\ell_2^2+\ell_3^2=\frac{1}{2} \end{aligned}$

$\begin{aligned} & \mathrm{AB}=5 \\ & \mathrm{AC}=5 \end{aligned}$

$\therefore \mathrm{D}$ is midpoint of $\mathrm{BC}$

$\begin{aligned} & \mathrm{D}\left(\frac{1}{2}, \frac{3}{2}, 3\right) \\ & \therefore l=\sqrt{\left(2-\frac{1}{2}\right)^2+\left(-3-\frac{3}{2}\right)^2+(3-3)^2} \\ & l=\sqrt{\frac{45}{2}} \\ & \therefore 2 l^2=45 \end{aligned}$

$\begin{aligned} & \vec{a} \cdot[(\vec{c} \times \vec{b})-\vec{b}-\vec{c}] \\ & \vec{a} \cdot(\vec{c} \times \vec{b})-\vec{a} \cdot \vec{b}-\vec{a} \cdot \vec{c} \quad \text{..... (i)} \end{aligned}$

$\begin{aligned} & \text { given } \vec{a} \times \vec{c}=\vec{b} \\ & \Rightarrow(\vec{a} \times \vec{c}) \cdot \vec{b}=\vec{b} \cdot \vec{b}=|\vec{b}|^2=27 \\ & \Rightarrow \vec{a} \cdot(\vec{c} \times \vec{b})=[\vec{a} \quad \vec{c} \quad \vec{b}]=(\vec{a} \times \vec{c}) \cdot \vec{b}=27 \quad \text{.... (ii)} \end{aligned}$

$\begin{aligned} & \text { Now } \vec{a} \cdot \vec{b}=3-6+3=0 \quad \text{.... (iii)}\\ & \vec{a} \cdot \vec{c}=3 \quad \text{.... (iv) (given)} \end{aligned}$

$\begin{gathered} \text { By (i), (ii), (iii) & (iv) } \\ 27-0-3=24 \end{gathered}$