(i) The vector equations of the line is:
\(\vec { r } =\vec { a } +\lambda (\vec { b } -\vec { a } )\)
\(\Rightarrow \vec { r } =(0\hat { i } +0\hat { j } +0\hat { k } )+\lambda ((5-0)\hat { i } +(-2-0)\hat { j } +(3-0)\hat { k } )\)
\(\Rightarrow \vec { r } =\lambda (5\hat { i } -2\hat { j } +3\hat { k } )\)
(ii) Cartesian equation is:
\(\frac { x-{ x }_{ 1 } }{ { x }_{ 2 }-{ x }_{ 1 } } =\frac { y-{ y }_{ 1 } }{ { y }_{ 2 }-{ y }_{ 1 } } =\frac { z-{ z }_{ 1 } }{ { z }_{ 2 }-{ z }_{ 1 } } \)
\(\Rightarrow \frac { x-0 }{ 5-0 } =\frac { y-0 }{ -2-0 } =\frac { z-0 }{ 3-0 } \)
\(\Rightarrow \frac { x }{ 5 } =\frac { y }{ -2 } =\frac { z }{ 3 } \)

The given lines are:
\(\vec{r}=(\hat{i}+2 \hat{j}+\hat{k})+\lambda(\hat{i}-\hat{j}+\hat{k})\)
\(\vec{r}=2 \hat{i}-\hat{j}-\hat{k}+\mu(2 \hat{i}+\hat{j}+2 \hat{k}) \)
It is known that the shortest distance between the lines, \(\vec{r}=\vec{a}_{1}+\lambda \vec{b}_{1} \text { and } \vec{r}=\vec{a}_{2}+\mu \vec{b}_{2}, \text { is } \)
\(\text { given by, }\)
\(d=\left|\frac{\left(\vec{b}_{1} \times \vec{b}_{2}\right) \cdot\left(\vec{a}_{2}-\vec{a}_{2}\right)}{\left|\vec{b}_{1} \times \vec{b}_{2}\right|}\right|\)
Comparing the given equations, we obtain \(\)
\(\vec{a}_{1}=\hat{i}+2 \hat{j}+\hat{k} \)
\(\vec{b}_{1}=\hat{i}-\hat{j}+\hat{k} \)
\(\vec{a}_{2}=2 \hat{i}-\hat{j}-\hat{k} \)
\(\vec{b}_{2}=2 \hat{i}+\hat{j}+2 \hat{k} \)
\(\vec{a}_{2}-\vec{a}_{1}=(2 \hat{i}-\hat{j}-\hat{k})-(\hat{i}+2 \hat{j}+\hat{k})=\hat{i}-3 \hat{j}-2 \hat{k} \)
\(\vec{b}_{1} \times \vec{b}_{2}=\left|\begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 1 \\ 2 & 1 & 2 \end{array}\right| \)
\(\vec{b}_{1} \times \vec{b}_{2}=(-2-1) \hat{i}-(2-2) \hat{j}+(1+2) \hat{k}=-3 \hat{i}+3 \hat{k} \)
\(\Rightarrow\left|\vec{b}_{1} \times \vec{b}_{2}\right|=\sqrt{(-3)^{2}+(3)^{2}}=\sqrt{9+9}=\sqrt{18}=3 \sqrt{2} \)
Substituting all the values in equation (1), we obtain
\(d=\left|\frac{(-3 \hat{i}+3 \hat{k}) \cdot(\hat{i}-3 \hat{j}-2 \hat{k})}{3 \sqrt{2}}\right| \)
\(\Rightarrow d=\left|\frac{-3.1+3(-2)}{3 \sqrt{2}}\right| \)
\(\Rightarrow d=\left|\frac{-9}{3 \sqrt{2}}\right| \)
\(\Rightarrow d=\frac{3}{\sqrt{2}}=\frac{3 \times \sqrt{2}}{\sqrt{2} \times \sqrt{2}}=\frac{3 \sqrt{2}}{2} \)
Therefore, the shortest distance between the two lines is \(\frac{3 \sqrt{2}}{2} \text { units. } \)

\(\frac{9}{\sqrt{171}}\) units.

Any plane parallel to:
\(\vec { r } =(\hat { i } +\hat { j } +\hat { k } )=2\)
is \(\vec { r } =(\hat { i } +\hat { j } +\hat { k } )=k\) ...(i)
This passes through the point(a,b,c)
i.e.  \((a\hat { i } +b\hat { j } +c\hat { k } )\)
 \(\therefore (a\hat { i } +b\hat { j } +c\hat { k } ).(\hat { i } +\hat { j } +\hat { k } )=k\)

\(\Rightarrow(a)(1)+(b)(1)+(c)(1)=k\)

\(\Rightarrow k=a+b+c.\)
putting in (1) the required equation of the plane is:
 \(\vec { r } .(\hat { i } +\hat { j } +\hat { k } )=a+b+c\)

\(i.e.(x\hat { i } +y\hat { j } +z\hat { k } ).(\hat { i } +\hat { j } +\hat { k } )=a+b+c\)

\(i.e x+y+z=a+b+c \)

Here 
\(\vec { r } =6\hat { i } +2\hat { j } +2\hat { k } +\lambda (\hat { i } -2\hat { j } +2\hat { k } ) \ and\)
\(\vec { r } =-4\hat { i } -\hat { k } +\mu (3\hat { i } -2\hat { j } -2\hat { k } )\)
\(\vec { { a }_{ 1 } } =6\hat { i } +2\hat { j } +2\hat { k } ,\)
\(\vec { { a }_{ 2 } } =-4\hat { i } -\hat { k } \)
\(and\quad \vec { { b }_{ 1 } } =\hat { i } -2\hat { j } +2\hat { k } ,\vec { { b }_{ 2 } } =3\hat { i } -2\hat { j } -2\hat { k } .\)
\(\therefore \vec { { a }_{ 1 } } -\vec { { a }_{ 2 } } =(6\hat { i } +2\hat { j } +2\hat { k } )-(-4\hat { i } -\hat { k } )\)
\(=10\hat { i } +2\hat { j } +3\hat { k } \)
\(and\quad \vec { { b }_{ 1 } } \times \vec { { b }_{ 2 } } \quad =\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 1 & -2 & 2 \\ 3 & -2 & -2 \end{matrix} \right| \)
\(=\hat { i } (4+4)-\hat { j } (-2-6)+\hat { k } (-2+6)\)
\(=8\hat { i } +8\hat { j } +4\hat { k } \)
\(\therefore \quad \left| \vec { { b }_{ 1 } } \times \vec { { b }_{ 2 } } \right| =\sqrt { 64+64+16 } \)
\(=\sqrt { 144 } =12\)
\(\text {S.D. between the given lines}\)
\(=\left| \frac { \left( \vec { { a }_{ 1 } } -\vec { { a }_{ 2 } } \right) .\left( \vec { { b }_{ 1 } } \times \vec { { b }_{ 2 } } \right) }{ \left| \vec { { b }_{ 1 } } \times \vec { { b }_{ 2 } } \right| } \right| =\left| \frac { (10\hat { i } +2\hat { j } +3\hat { k } ).(8\hat { i } +8\hat { j } +4\hat { k } ) }{ 12 } \right| \)
\(=\left| \frac { (10)(8)+(2)(8)+(3)(4) }{ 12 } \right| =\left| \frac { 80+16+12 }{ 12 } \right| =\left| \frac { 108 }{ 12 } \right| =9units.\)

The given planes are
\(\vec{r} \cdot(\hat{i}+\hat{j}+\hat{k})=1 \)
\(\Rightarrow \vec{r} \cdot(\hat{i}+\hat{j}+\hat{k})-1=0 \)
\(\vec{r} \cdot(2 \hat{i}+3 \hat{j}-\hat{k})+4=0 \)
The equation of any plane passing through the line of intersection of these planes is
\({[\vec{r} \cdot(\hat{i}+\hat{j}+\hat{k})-1]+\lambda[\vec{r} \cdot(2 \hat{i}+3 \hat{j}-\hat{k})+4]=0} \)
\(\vec{r} \cdot[(2 \lambda+1) \hat{i}+(3 \lambda+1) \hat{j}+(1-\lambda) \hat{k}]+(4 \lambda+1)=0 \)
Its direction ratios are \( (2 \lambda+1),(3 \lambda+1), \text { and }(1-\lambda) \text {. }\)
The required plane is parallel to  x -axis. Therefore, its normal is perpendicular to } x -axis.
The direction ratios of x -axis are 1, 0, and 0 .
\(\therefore 1 .(2 \lambda+1)+0(3 \lambda+1)+0(1-\lambda)=0 \)
\(\Rightarrow 2 \lambda+1=0 \)
\(\Rightarrow \lambda=-\frac{1}{2} \)
Substituting \( lambda=-\frac{1}{2} \) in equation (1), we obtain
\(\Rightarrow \vec{r} \cdot\left[-\frac{1}{2} \hat{j}+\frac{3}{2} \hat{k}\right]+(-3)=0 \)
\(\Rightarrow \vec{r}(\hat{j}-3 \hat{k})+6=0 \)
Therefore, its Cartesian equation is y - 3z + 6 = 0
This is the equation of the required plane.

The given line is 
\(\vec { r } =\left( 2\hat { i } -\hat { j } +2\hat { k } \right) +\lambda \left( 3\hat { i } +4\hat { j } +2\hat { k } \right)....(i)\)
and  the given plane is \( \vec { r } .\left( \hat { i } -\hat { j } +\hat { k } \right) =5 ....(ii)\)
Solving (i) and (ii)
\([2\hat { i } -\hat { j } +2\hat { k } +\lambda \left( 3\hat { i } +4\hat { j } +2\hat { k } \right) ].\left( \hat { i } -\hat { j } +\hat { k } \right) =5\)
\( \Rightarrow (2+3\lambda )\hat { i } +(4\lambda -1)\hat { j } +(2\lambda +2)\hat { k } ).\left( \hat { i } -\hat { j } +\hat { k } \right) =\lambda\)
\(\Rightarrow (2+3\lambda )(1)+(4\lambda -1)(-1)+(2\lambda +2)(1)=5\)
\(\Rightarrow 2+3\lambda -4\lambda +1+2\lambda +2=5\)
\(\Rightarrow \lambda =0\)
the point of intersection of(i) are (ii) is
\(\left( 2\hat { i } -\hat { j } +2\hat { k } \right) i.e.(2,-1,2).\)
The other point is \((-1,-5,-10)\)
Required distance \(=\sqrt { { \left( -1-2 \right) }^{ 2 }+{ \left( -5+1 \right) }^{ 2 }+{ \left( -10-2 \right) }^{ 2 } } =\sqrt { 9+16+144 } =13\)

The line parallel to x-­­axis and passing through the origin is x-axis itself.
Let A be a point on x-axis. Therefore, the coordinates of A are given by (a, 0, 0), where a ∈ R.
Direction ratios of OA are (a − 0) = a, 0, 0
The equation of OA is given by,
\(\frac{x-0}{a}=\frac{y-0}{0}=\frac{z-0}{0} \)
\(\Rightarrow \frac{x}{1}=\frac{y}{0}=\frac{z}{0}=a \)
Thus, the equation of line parallel to x-axis and passing through origin is \(\)
\(\frac{x}{1}=\frac{y}{0}=\frac{z}{0}\)

The given equation of the plane is:
2x + y - z = 5
\(\Rightarrow \frac { x }{ 5/2 } +\frac { y }{ 5 } +\frac { z }{ -5 } =1\)
Comparing with 
\(\frac { x }{ a } +\frac { y }{ b } +\frac { z }{ c } =1\)
the intercepts on the axes are 5/2, 5 and -5.

\(\text { Here, } \vec{a}=2 \hat{i}+5 \hat{j}-3 \hat{k}, \overrightarrow{\mathrm{N}}=6 \hat{i}-3 \hat{j}+2 \hat{k} \text { and } d=4\)
Therefore, the distance of the point (2, 5, – 3) from the given plane is
\(\frac{|(2 \hat{i}+5 \hat{j}-3 \hat{k}) \cdot(6 \hat{i}-3 \hat{j}+2 \hat{k})-4|}{|6 \hat{i}-3 \hat{j}+2 \hat{k}|}=\frac{|12-15-6-4|}{\sqrt{36+9+4}}=\frac{13}{7}\)

The given equation can be written as
\(\vec{r} \cdot(-6 \hat{i}+3 \hat{j}+2 \hat{k})=1\)
\(\text {Now } \ |-6 \hat{i}+3 \hat{j}+2 \hat{k}|=\sqrt{36+9+4}=7\)
Therefore, dividing both sides of (1) by 7, we get
\(\vec{r} \cdot\left(-\frac{6}{7} \hat{i}+\frac{3}{7} \hat{j}+\frac{2}{7} \hat{k}\right)=\frac{1}{7}\)
which is the equation of the plane in the form \( \vec{r} \cdot \hat{n}=d \text { . }\)
This shows that \( \hat{n}=-\frac{6}{7} \hat{i}+\frac{3}{7} \hat{j}+\frac{2}{7} \hat{k} \)is a unit vector perpendicular to the plane through the origin. Hence, the direction cosines of  \( \hat{n} \text { are } \frac{-6}{7}, \frac{3}{7}, \frac{2}{7} \text { . }\)

The x-axis makes angles 0°, 90° and 90° respectively with x, y and z-axis.
Therefore, the direction cosines of x-axis are cos 0°, cos 90°, cos 90° i.e., 1, 0, 0.
Similarly, direction cosines of y-axis and z-axis are 0, 1, 0 and 0, 0, 1 respectively.

We know that the direction-cosines of the line joining P(x₁, y₁, z₁) and q(x₂, y₂, z₂) are:
\(\frac{x_{2}-x_{1}}{\mathrm{PQ}}, \frac{y_{2}-y_{1}}{\mathrm{PQ}}, \frac{z_{2}-z_{1}}{\mathrm{PQ}} \)
\(\text {where } \mathrm{PQ}=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}+\left(z_{2}-z_{1}\right)^{2}} \)
Here P is (– 2, 4, – 5) and Q is (1, 2, 3).
\(\mathrm{PQ}=\sqrt{(1-(-2))^{2}+(2-4)^{2}+(3-(-5))^{2}}=\sqrt{77}\)
Thus, the direction cosines of the line joining two points is
\(\frac{3}{\sqrt{77}}, \frac{-2}{\sqrt{77}}, \frac{8}{\sqrt{77}}\)

The line has direction -ratios <2,-1,-2>,determine its direction cosines are:
\(<\frac { 2 }{ \sqrt { 4+1+4 } } ,\frac { -1 }{ \sqrt { 4+1+4 } } ,\frac { -2 }{ \sqrt { 4+1+4 } } >\)
\(i.e. <\frac { 2 }{ 3 } ,\frac { -1 }{ 3 } ,\frac { -2 }{ 3 } >\)