Taking \(LHS=\overrightarrow a.\{(\overrightarrow b+\overrightarrow c)\times\overrightarrow d\}\)
\(=\overrightarrow a.\{(\overrightarrow b \times\overrightarrow d)+(\overrightarrow c\times\overrightarrow d)\}\)
\(=\overrightarrow a.(\overrightarrow b\times \overrightarrow d)+\overrightarrow a.(\overrightarrow c\times \overrightarrow d)\)
\(=\left[ \begin{matrix} \overrightarrow { a } , & \overrightarrow { b } , & \overrightarrow { d } \end{matrix} \right] +\left[ \begin{matrix} \overrightarrow { a } , & \overrightarrow { c } , & \overrightarrow { d } \end{matrix} \right] \)

\(\overrightarrow a+\overrightarrow b+\overrightarrow c=0, \therefore \overrightarrow a+\overrightarrow b=-\overrightarrow c\)
\(\Rightarrow (\overrightarrow a+\overrightarrow b)^{ 2 }=(-\overrightarrow c)^{ 2}=\left| \overrightarrow {c } \right| ^{2 }\)
\(\Rightarrow9+25+2\left| \overrightarrow { a } \right| \left| \overrightarrow { b } \right| \cos\theta=49\)
\(\theta\) being angle between \(\overrightarrow a\) and \(\overrightarrow b,\)
\(\therefore \cos\theta=\frac{15}{2.3.5}=\frac{1}{2}\Rightarrow\theta=\frac{\pi}{3}\)

The position vector of point R dividing the line segment joining two points P and Q in the ratio m : n is given by: 
 i. Internally: 
\(\frac{m \vec{b}+n \vec{a}}{m+n}\)
 ii. Externally:  \(\)
\(\frac{m \vec{b}-n \vec{a}}{m-n}\)
Position vectors of  P and Q are given as: \(\)
\(\overrightarrow{\mathrm{OP}}=\hat{i}+2 \hat{j}-\hat{k} \text { and } \overrightarrow{\mathrm{OQ}}=-\hat{i}+\hat{j}+\hat{k}\)
(i) The position vector of point R which divides the line joining two points P and Q internally in the ratio 2: 1 is given by, 
\(\overrightarrow{O R} =\frac{2(-\hat{i}+\hat{j}+\hat{k})+1(\hat{i}+2 \hat{j}-\hat{k})}{2+1}=\frac{(-2 \hat{i}+2 \hat{j}+2 \hat{k})+(\hat{i}+2 \hat{j}-\hat{k})}{3} \)
\(=\frac{-\hat{i}+4 \hat{j}+\hat{k}}{3}=-\frac{1}{3} \hat{i}+\frac{4}{3} \hat{j}+\frac{1}{3} \hat{k} \)
(ii) The position vector of point R which divides the line joining two points P and Q externally in the ratio 2 : 1 is given by,
\(\overrightarrow{\mathrm{OR}} =\frac{2(-\hat{i}+\hat{j}+\hat{k})-1(\hat{i}+2 \hat{j}-\hat{k})}{2-1}=(-2 \hat{i}+2 \hat{j}+2 \hat{k})-(\hat{i}+2 \hat{j}-\hat{k}) \)
\(=-3 \hat{i}+3 \hat{k} \)

\( =\left|\begin{array}{rrr} 2 & -3 & 1 \\ 1 & -1 & 2 \\ 2 & 1 & -1 \end{array}\right| \)
\(=2(-1)+3(-5)+1(3) \)
\(=|-2-15+3|=|-14|=14 \mathrm{cu} \text { units } \)

\((\overrightarrow { a } -\overrightarrow { d } )\) x \((\overrightarrow { b } -\overrightarrow { c } )\) 
\(=\overrightarrow { a } *\overrightarrow { b } -\overrightarrow { a } *\overrightarrow { c } -\overrightarrow { d } *\overrightarrow { b } +\overrightarrow { d } *\overrightarrow { c } \)
\(=\overrightarrow { c } *\overrightarrow { d } -\overrightarrow { b } *\overrightarrow { d } -\overrightarrow { d } *\overrightarrow { b } +\overrightarrow { d } -\overrightarrow { c } \)
\([\because \overrightarrow { a } *\overrightarrow { b } =\overrightarrow { c } -\overrightarrow { d } and\overrightarrow { a } *\overrightarrow { c } =\overrightarrow { b } *\overrightarrow { d } ]\)
\(=\overrightarrow { c } *\overrightarrow { d } -\overrightarrow { b } *\overrightarrow { d } +\overrightarrow { b } *\overrightarrow { d } -\overrightarrow { c } *\overrightarrow { d } =\overrightarrow { 0 } \) 
Hence, \((\overrightarrow { a } -\overrightarrow { d } )\) is parallel to \((\overrightarrow { b } -\overrightarrow { c } )\)

\(\vec{r} =2 \vec{a}-\vec{b}+3 \vec{c} \)
\(=2 \hat{i}+2 \hat{j}+2 \hat{k}-4 \hat{i}+2 \hat{j}-3 \hat{k}+3 \hat{i}-6 \hat{j}+3 \hat{k} \)
\(\Rightarrow \vec{r}=\hat{i}-2 \hat{j}+2 \hat{k} \)
Vector of magnitude 6 units and parallel to
\((2 \hat a-\hat b+3\hat c) \text { is } 6 \hat{r}\)
\(\text { Vector }=6\left(\frac{\hat{i}-2 \hat{j}+2 \hat{k}}{\sqrt{1+4+4}}\right)=2 \hat{i}-4 \hat{j}+4 \hat{k}\)

\(\vec{b}+\vec{c}=3 \hat{i}+\hat{j}+2 \hat{k},\)
\(\text { Projection of }(\vec{b}+\vec{c}) \text { on } \vec{a}=\frac{(\vec{b}+\vec{c}) \cdot \vec{a}}{|\vec{a}|} \text { , }\)
\(=\frac{(3 \hat{i}+\hat{j}+2 \hat{k}) \cdot(2 \hat{i}-2 \hat{j}+\hat{k})}{\sqrt{4+4+1}}=\frac{6-2+2}{3}=\frac{6}{3}=2\)