\(\overrightarrow { a } =\hat { i } -\hat { j } +7\hat { k } \) and \(\overrightarrow { b } =5\hat { i } -\hat { j } +\lambda \hat { k } \) 
Then  \(\overrightarrow { a } +\overrightarrow { b } =(\hat { i } -\hat { j } +7\hat { k } )+(5\hat { i } -\hat { j } +\lambda \hat { k } )\) 
\(=6\hat { i } -2\hat { j } +(7+\lambda )\hat { k } \) 
and  \(\overrightarrow { a } -\overrightarrow { b } =(\hat { i } -\hat { j } +7\hat { k } )-(5\hat { i } -\hat { j } +\lambda \hat { k } )\) 
\(=-4\hat { i } +(7-\lambda )\hat { k } \) 
Now \((\overrightarrow { a } +\overrightarrow { b } )\) and \((\overrightarrow { a } -\overrightarrow { b } )\)  are perpendicular vectors.
\(\therefore (\overrightarrow { a } +\overrightarrow { b } ).(\overrightarrow { a } -\overrightarrow { b } )=0\)
\(\Rightarrow (6\hat { i } -2\hat { j } +(7+\lambda )\hat { k } ).(-4\hat { i } +0\hat { j } +(7-\lambda )\hat { k } )=0\)
\(\Rightarrow (6)(-4)+(-2)(0)+(7+\lambda )+(7-\lambda )=0\)
\(\Rightarrow -24+0+498-{ \lambda }^{ 2 }=0\Rightarrow { \lambda }^{ 2 }=25\)
Hence \(\lambda =\pm 5\) 

m =cos\(\pi\over4\) = \(\frac { 1 }{ \sqrt { 2 } } \) and  n = cos\(\pi\over2\)=0
Now \({ i }^{ 2 }+{ m }^{ 2 }+{ n }^{ 2 }=1\) gives \({ i }^{ 2 }+\frac { 1 }{ 2 } +0=1\)
\(\Rightarrow { i }^{ 2 }=\frac { 1 }{ 2 } \quad \Rightarrow i=\pm \frac { 1 }{ \sqrt { 2 } } \) 
Hence, the reqd.vector is \(\overset { \wedge }{ r } =3\sqrt { 2 } (l\overset { \wedge }{ i } +m\overset { \wedge }{ j } +n\overset { \wedge }{ k } )\) 
\(i.e\quad \overset { \rightarrow }{ r } =3\sqrt { 2 } \left( \pm \frac { 1 }{ \sqrt { 2 } } \overset { \wedge }{ i } +\frac { 1 }{ \sqrt { 2 } } \overset { \wedge }{ j } +0\overset { \wedge }{ k } \right) i.e\overset { \rightarrow }{ r } =\pm 3\overset { \wedge }{ i } +3\overset { \wedge }{ j } \)

let  \(\overset { \rightarrow }{ a } ={ a }_{ 1 }\overset { \wedge }{ i } +{ a }_{ 2 }\overset { \wedge }{ j } +{ a }_{ 3 }\overset { \wedge }{ k } \) 
\(\overset { \rightarrow }{ b } ={ b }_{ 1 }\overset { \wedge }{ i } +{ b }_{ 2 }\overset { \wedge }{ j } +{ b }_{ 3 }\overset { \wedge }{ k } \) 
and   \(\overset { \rightarrow }{ c } ={ c }_{ 1 }\overset { \wedge }{ i } +{ c }_{ 2 }\overset { \wedge }{ j } +{ c }_{ 3 }\overset { \wedge }{ k } \) 
Since  \(\overset { \rightarrow }{ a } \pm \overset { \rightarrow }{ b } \) 
\(\therefore \)  \({ a }_{ 1 }{ b }_{ 1 }+{ a }_{ 2 }{ b }_{ 2 }+{ a }_{ 3 }{ b }_{ 3 }=0\)...(1)
Similarly \({ b }_{ 1 }{ c }_{ 1 }+{ b }_{ 2 }{ c }_{ 2 }+{ b }_{ 3 }{ c }_{ 3 }=0\) ...(2)
and   \({ a }_{ 1 }{ c }_{ 1 }+{ a }_{ 2 }{ c }_{ 2 }+{ a }_{ 3 }{ c }_{ 3 }=0\)  ..(3)
Now  \({ \left[ \overset { \rightarrow }{ a } .\left( \overset { \rightarrow }{ b } \times \overset { \rightarrow }{ c } \right) \right] }^{ 2 }={ \left[ \overset { \rightarrow }{ a } \overset { \rightarrow }{ b } \overset { \rightarrow }{ c } \right] }^{ 2 }\)
\(=\left| \begin{matrix} { a }_{ 1 } & { a }_{ 2 } & { a }_{ 3 } \\ { b }_{ 1 } & { b }_{ 2 } & { b }_{ 3 } \\ { c }_{ 1 } & { c }_{ 2 } & { c }_{ 3 } \end{matrix} \right| \left| \begin{matrix} { a }_{ 1 } & { a }_{ 2 } & { a }_{ 3 } \\ { b }_{ 1 } & { b }_{ 2 } & { b }_{ 3 } \\ { c }_{ 1 } & { c }_{ 2 } & { c }_{ 3 } \end{matrix} \right| \) 
\(=\left| \begin{matrix} { a }_{ 1 } & { a }_{ 2 } & { a }_{ 3 } \\ { b }_{ 1 } & { b }_{ 2 } & { b }_{ 3 } \\ { c }_{ 1 } & { c }_{ 2 } & { c }_{ 3 } \end{matrix} \right| \left| \begin{matrix} { a }_{ 1 } & { a }_{ 2 } & { a }_{ 3 } \\ { b }_{ 1 } & { b }_{ 2 } & { b }_{ 3 } \\ { c }_{ 1 } & { c }_{ 2 } & { c }_{ 3 } \end{matrix} \right| \) 
\(=\left| \begin{matrix} { { a }_{ 1 } }^{ 2 } & { { a }_{ 2 } }^{ 2 } & { { a }_{ 3 } }^{ 2 } \\ { a }_{ 1 }{ b }_{ 1 } & { a }_{ 2 }{ b }_{ 2 } & { a }_{ 3 }{ b }_{ 3 } \\ { a }_{ 1 }{ c }_{ 1 } & { a }_{ 2 }{ c }_{ 2 } & { a }_{ 3 }{ c }_{ 3 } \end{matrix}\begin{matrix} { a }_{ 1 }{ b }_{ 1 } & { a }_{ 2 }{ b }_{ 2 } & { a }_{ 3 }{ b }_{ 3 } \\ { b_{ 1 } }^{ 2 } & { { b }_{ 2 } }^{ 2 } & { { b }_{ 3 } }^{ 2 } \\ { b }_{ 1 }{ c }_{ 1 } & { b }_{ 2 }{ c }_{ 2 } & { b }_{ 3 }{ c }_{ 3 } \end{matrix}\begin{matrix} { a }_{ 1 }{ c }_{ 1 } & { a }_{ 2 }{ c }_{ 2 } & { a }_{ 3 }{ c }_{ 3 } \\ { b }_{ 1 }{ c }_{ 1 } & { b }_{ 2 }{ c }_{ 2 } & { b }_{ 3 }{ c }_{ 3 } \\ { c_{ 1 } }^{ 2 } & { { c }_{ 2 } }^{ 2 } & { { c }_{ 3 } }^{ 2 } \end{matrix} \right| \) 
\(=\left| \begin{matrix} { a }^{ 2 } & 0 & 0 \\ 0 & { b }^{ 2 } & 0 \\ 0 & 0 & { c }^{ 2 } \end{matrix} \right| ={ a }^{ 2 }{ b }^{ 2 }{ c }^{ 2 }\)
\(\left[ using(1)(2)(3)\& { a }^{ 2 }={ { a }_{ 1 } }^{ 2 }+{ { a }_{ 2 } }^{ 2 }+{ { a }_{ 3 } }^{ 2 };etc. \right] \) 

Given,\(|\vec{r}|=14\) and if \(\vec{r}=a \hat{i}+b \hat{j}+c \hat{k}\) then \(a=2 \lambda, b=3 \lambda\)  \(\text { and } c=-6 \lambda \text { for some } \lambda \neq 0 \text { . }\)
\(\therefore \text { Direction cosines } l, m \text { and } n \text { are } l=\frac{a}{|\vec{r}|}=\frac{2 \lambda}{14}=\frac{\lambda}{7}\)
\(m=\frac{b}{|\vec{r}|}=\frac{3 \lambda}{14} \text { and } n=\frac{c}{|\vec{r}|}=\frac{-6 \lambda}{14}=\frac{-3 \lambda}{7}\) 
Also, we know that \(l^{2}+m^{2}+n^{2}=1\) 
\( \therefore \frac{\lambda^{2}}{49}+\frac{9 \lambda^{2}}{196}+\frac{9 \lambda^{2}}{49}=1 \)
\(\Rightarrow \frac{4 \lambda^{2}+9 \lambda^{2}+36 \lambda^{2}}{196}=1 \)
\(\Rightarrow 49 \lambda^{2}=196 \Rightarrow \lambda^{2}=\frac{196}{49} \)
\(\lambda^{2}=4 \Rightarrow \lambda=\pm 2 \) 
So, the direction cosines l,m and n are \(\frac{2}{7}, \frac{3}{7} \text { and } \frac{-6}{7}\) .
[\(\because \vec{r} \)makes an acute angle with X -axis, so we will take positive value of \(\lambda\)]
\(\because \ \vec{r}=\hat{r} \cdot|\vec{r}|\) 
\(\therefore \vec{r}=(l \hat{i}+m \hat{j}+n \hat{k}) \cdot|\vec{r}|=\left(\frac{2}{7} \hat{i}+\frac{3}{7} \hat{j}-\frac{6}{7} \hat{k}\right) \cdot 14\) 
\(=4 \hat{i}+6 \hat{j}-12 \hat{k}\) 
 Thus, the components of \(\vec{r} \text { are } 4 \hat{i}, 6 \hat{j} \text { and }-12 \hat{k}\).

Clearly, \(\vec{b}+\vec{c}=\left(b_{1} \hat{i}+b_{2} \hat{j}+b_{3} \hat{k}\right)+\left(c_{1} \hat{i}+c_{2} \hat{j}+c_{3} \hat{k}\right)\) 
\(=\left(b_{1}+c_{1}\right) \hat{i}+\left(b_{2}+c_{2}\right) \hat{j}+\left(b_{3}+c_{3}\right) \hat{k}\) 
\(\therefore \vec{a} \times(\vec{b}+\vec{c})=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ a_{1} & a_{2} & a_{3} \\ b_{1}+c_{1} & b_{2}+c_{2} & b_{3}+c_{3} \end{array}\right|\) 
\(=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ a_{1} & a_{2} & a_{3} \\ b_{1} & b_{2} & b_{3} \end{array}\right|+\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ a_{1} & a_{2} & a_{3} \\ c_{1} & c_{2} & c_{3} \end{array}\right|\) 
[by property of determinant]
\(=(\vec{a} \times \vec{b})+(\vec{a} \times \vec{c})\) 
Hence, \(\vec{a} \times(\vec{b}+\vec{c})=(\vec{a} \times \vec{b})+(\vec{a} \times \vec{c})\)