The given cartesian equation of the line is:
\(\frac { x-5 }{ 3 } =\frac { y+4 }{ 7 } =\frac { z-6 }{ 2 } \) . (1) 
This passes through the point(5,-4,6) and has <3,7,2> as its direction ratios
\(\Rightarrow\) the line (1) passes through the point A\((\vec { a } )\)where
\(\vec { a } =5\vec { i } -4\vec { j } +6\vec { k } \ \text {and is in the direction of}\)
\( \vec { m } =3\vec { i } +7\vec { j } +2\vec { k } \)
The vector equation of the required line is:
\(\vec { r } =\vec { a } +\lambda \vec { m } \)
\(i.e.\ \vec { r } =\left( 5\vec { i } -4\vec { j } +6\vec { k } \right) +\lambda \left( \vec { i } +7\vec { j } +2\vec { k } \right) \)

It is given that the line passes through the point A (1, 2, 3). Therefore, the position vector through A 
\(\vec { a } =\hat { i } +2\hat { j } +3\hat { k } \quad and\)
\(\vec { b } =3\vec { i } +2\vec { j } -2\vec { k } \)
The equation of the required line is:
\(\vec { r } =\left( \hat { i } +2\hat { j } +3\hat { k } \right) +\lambda \left( 3\vec { i } +2\vec { j } -2\vec { k } \right) ;\lambda \in R\)

Let A(2, 3, 4); B(-1, -2, 1) and C(5, 8, 7) be the given points
Direction ratios of AB are: (-1-2, -2-3, 1-4)
i.e. (-3, -5, -3) i.e. (3, 5, 3)
Direction-ratios of BC are: (5-(-1), 8-(-2(,7-1)
i.e. (6, 10, 6) i.e. (3, 5, 3)
\(\Rightarrow\) AB and BC have same direction ratios
\(\Rightarrow\) AB||BC
But B is a common point.Hence A, B, C are collinear.

The direction ratios of the first line are 3, 5, 4 and the direction ratios of the second line are 1, 1, 2. If θ is the angle between them, then
\(cos\theta =\left| \frac { (3)(1)+(5)(1)+(4)(2) }{ \sqrt { 9+25+16 } \sqrt { 1+1+4 } } \right| =\left| \frac { 3+5+8 }{ \sqrt { 50 } \sqrt { 6 } } \right| =\frac { 16 }{ 5\sqrt { 2 } \sqrt { 6 } } =\frac { 8\sqrt { 3 } }{ 15 } \)
Hence,the required angle is \(\theta -\cos ^{ -1 }{ \left( \frac { 8\sqrt { 3 } }{ 15 } \right) } \)

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 given lines are:
 \(\vec { r } =\hat { i } +\hat { j } +\lambda (2\hat { i } -\hat { j } +\hat { k } )\quad and\)
\(\vec { r } =2\hat { i } +\hat { j } -\hat { k } +\mu (3\hat { i } -5\hat { j } +2\hat { k } )\)
Comparing with \(\vec { r } =\vec { { a }_{ 1 } } +\lambda \vec { b } and\quad \vec { r } =\vec { { a }_{ 2 } } +\lambda \vec { b } ,\text {we have}:\)
\(\vec { r } =\vec { { a }_{ 1 } } +\lambda \vec { b } \ and\quad \vec { r } =\vec { { a }_{ 2 } } +\lambda \vec { b } ,we\quad have:\)
\(\vec { { b }_{ 1 } } =2\hat { i } -\hat { j } +\hat { k } \ and\quad \vec { { b }_{ 2 } } =3\hat { i } -5\hat { j } +2\hat { k } ;\)
\(\vec { { a }_{ 1 } } =\hat { i } +\hat { j } \ and \ \vec { { a }_{ 2 } } =2\hat { i } +\hat { j } -\hat { k } \)
\( \therefore \vec{a}_2-\vec{a}_1=\hat{i}-\hat{k} \\ \vec{b}_1 \times \vec{b}_2=(2 \hat{i}-\hat{j}+\hat{k}) \times(3 \hat{i}-5 \hat{j}+2 \hat{k}) \)
\(\vec { { b }_{ 1 } } \times \vec { { b }_{ 2 } } =\left| \begin{matrix} \hat { i } & \hat { j } & \hat { k } \\ 2 & -1 & 1 \\ 3 & -5 & 2 \end{matrix} \right| =3\hat { i } -\hat { j } -7\hat { k } \)
\(\left|\vec{b}_{1} \times \vec{b}_{2}\right|=\sqrt{9+1+49}=\sqrt{59}\)
Hence, the shortest distance between the given lines is given by
\(d=\left|\frac{\left(\vec{b}_{1} \times \vec{b}_{2}\right) \cdot\left(\vec{a}_{2}-\vec{a}_{1}\right)}{\left|\vec{b}_{1} \times \vec{b}_{2}\right|}\right|=\frac{|3-0+7|}{\sqrt{59}}=\frac{10}{\sqrt{59}}\)

Consider, the line \( \vec{r}=(1+\lambda) \hat{i}+(2-\lambda) \hat{j}+(\lambda+1) \hat{k} \)
\(i.e., \vec{r}=(\hat{i}+2 \hat{j}+\hat{k})+\lambda(\hat{i}-\hat{j}+\hat{k}) \)
\(\text {Here } \vec{a}_{1}=\hat{i}+2 \hat{j}+\hat{k}, \vec{b}_{1}=\hat{i}-\hat{j}+\hat{k} \)
Shortest distance between the lines 
\(=\left|\frac{\left(\overrightarrow{a_{2}}-\vec{a}_{1}\right) \cdot\left(\vec{b}_{1} \times \overrightarrow{b_{2}}\right)}{\left|\vec{b}_{1} \times \overrightarrow{b_{2}}\right|}\right|\)
\(\text {Now } \vec{a}_{2}-\vec{a}_{1}=2 \hat{i}-\hat{j}-\hat{k}-\hat{i}-2 \hat{j}-\hat{k}=\hat{i}-3 \hat{j}-2 \hat{k}\)
\(\vec{b}_{1} \times \vec{b}_{2} =\left|\begin{array}{rrr} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & 1 \\ 2 & 1 & 2 \end{array}\right| \)
\(-\hat{i}(-2-1)-\hat{j}(2-2)+\hat{k}(1+2) \)
\(=-3 \hat{i}+3 \hat{k} \)
\(\left|\vec{b}_{1} \times \vec{b}_{2}\right| =\sqrt{9+9}=3 \sqrt{2} \)
Shortest distance = \(\left| \frac { (\acute { i } -3\acute { j } -2\acute { k } ).(-3\acute { i } +3\acute { k } ) }{ 3\sqrt { 2 } } \right| =\left| \frac { -3-6 }{ 3\sqrt { 2 } } \right| =\frac { 3 }{ \sqrt { 2 } } =\frac { 3\sqrt { 2 } }{ 2 } \) units

General point on the line \( \frac{x-1}{3}=\frac{y}{2}=\frac{z+1}{7} \text { is }\)
\(Q(3 \lambda+1,2 \lambda, 7 \lambda-1)\)
If this point lies on the plane then
\(3 \lambda+1+2 \lambda-7 \lambda+1=8 \)
\(\Rightarrow-2 \lambda=6 \Rightarrow \lambda=-3 \)
Substituting in (i), point of intersection is
\(Q(-8,-6,-22)\)
Direction ratios of  P Q are12,12, 24 or 1, 1, 2
Equation of PQ is \(\frac { x-4 }{ 1 } =\frac { y-6 }{ 1 } =\frac { z-2 }{ 2 } \)

General point on the line
\(\frac{x-11}{10}=\frac{y+2}{-4}=\frac{z+8}{-11}=\lambda \text { (say) is } \)
\(Q(10 \lambda+11,-4 \lambda-2,-11 \lambda-8) \)
Direction ratios of PQ are
\(10 \lambda+11-2,-4 \lambda-2+1,-11 \lambda-8-5 \)
\(\text { i.c. } 10 \lambda+9,-4 \lambda-1,-11 \lambda-13 \)
 If PQ is perpendicular to the given line, then 
\(10(10 \lambda+9)-4(-4 \lambda-1)-11(-11 \lambda-13)=0 \)
\(\Rightarrow 237 \lambda=-237 \Rightarrow \lambda=-1 \)
Substituting in (i), we get the foot of perpendicular as Q(1, 2, 3)
Length of perpendicular
\(P Q =\sqrt{(2-1)^{2}+(-1-2)^{2}+(5-3)^{2}} \)
\(=\sqrt{1+9+4}=\sqrt{14} . \)

Equation of line through (3,- 4,- 5) and (2,- 3,1)
\({ { x-3}\over{2-3 } }={ { y-(-4)}\over{-3-(-4) } }={ { z-(-5)}\over{ 1-(-5)} }\)
\(\Rightarrow\) \({ {x-3 }\over{ -1} }={ {y+4 }\over{1 } }={ { z+5}\over{ 6} }=\lambda\)(Let)
\(\therefore\) Co-ordinate of any random point on this line
M \((-\lambda+3,\lambda-4,6\lambda-5).\)
Also, equation of plane passing through the points (2, 2,1), (3, 0,1) and (4, -1, 0) is given by the formula:
\(\begin{vmatrix} x-{ x }_{ 1 } & y-{ y }_{ 1 } & z-{ z }_{ 1 } \\ { x }_{ 2 }-{ x }_{ 1 } & { y }_{ 2 }-{ y }_{ 1 } & { z }_{ 2 }-{ z }_{ 1 } \\ { x }_{ 3 }-{ x }_{ 1 } & { y }_{ 3 }-{ y }_{ 1 } & { z }_{ 3 }-{ z }_{ 1 } \end{vmatrix}=0\)
i.e.,  \(\begin{vmatrix} x-2 & y-2 & z-1 \\ 3-2 & 0-2 & 1-1 \\ 4-2 & -1-2 & 0-1 \end{vmatrix}=0\)
\(\Rightarrow\)   \(\begin{vmatrix} x-2 & y-2 & z-1 \\ 1 & -2 & 0 \\ 2 & -3 & -1 \end{vmatrix}=0\)
\(\Rightarrow\)  ( x - 2 )( 2 - 0 ) - ( y - 2 ) ( - 1 - 0 )+ ( z - 1 ) ( - 3 + 4 ) = 0
\(\Rightarrow\) 2x + y + z  -7 = 0 ..(i)
When the line crosses the plane, M must satisfy the plane (i).
So, we have
\(2(-\lambda+3)+(\lambda-4)+(6\lambda-5)-7=0\)
\(\Rightarrow\)  \(\lambda=2\)
Hence, required co-ordinates of the point of intersection is M (1, - 2, 7).

\(\therefore\) Co-ordinates of Pare (1, - 2, 7).
Now, for ratio:
Let P divides AB in the ratio k : 1.
\(\therefore\) Using section formula
\(1={{2k+3}\over{k+1}}\)
\(\Rightarrow\) 2k + 3 = k + 1
\(\therefore\)  k = -2
Ratio is - 2 : 1 or 2 : 1 externally.

Equation of plane passing through the intersection of two given planes is
\(\vec { r } .\left( \overset { \wedge }{ i } +\overset { \wedge }{ j } +4\overset { \wedge }{ k } \right) -1+\lambda \left\{ \vec { r } .(2\hat { i } +3\hat { j } -\hat { k } )+4 \right\} =0\)
\(\vec { r } .\left\{ (1+2\lambda )\hat { i } +(1+3\lambda )\hat { j } +(1-\lambda )\hat { k } \right\} -1+4\lambda \)
Plane (i) is to the plane 
∴ 1(1+2λ) + 1(1+3λ) - 4(1-λ) = 0
⇒ 9λ = 2 ⇒ \(\lambda =\frac { 2 }{ 9 } \)
∴Eqn.of plane is \(\vec { r } .\left( \frac { 13 }{ 9 } \hat { i } +\frac { 15 }{ 9 } \hat { j } +\frac { 7 }{ 9 } \hat { k } \right) -\frac { 1 }{ 9 } =0\)
or 13x + 15 y +7z -1 = 0

Lines \({{x-{x}_{1}}\over{{l}_{1}}}={{y-{y}_{1}}\over{{m}_{1}}}={{z-{z}_{1}}\over{{n}_{1}}}\) and \({{x-{x}_{2}}\over{{l}_{2}}}={{y-{y}_{2}}\over{{m}_{2}}}={{z-{z}_{2}}\over{{n}_{2}}}\) are coplaner, if
\(\begin{vmatrix} { x }_{ 2 }-{ x }_{ 1 } & { y }_{ 2 }-{ y }_{ 1 } & { z }_{ 2 }-{ z }_{ 1 } \\ { l }_{ 1 } & { m }_{ 1 } & { n }_{ 1 } \\ { l }_{ 2 } & { m }_{ 2 } & { n }_{ 2 } \end{vmatrix}=0\)
Now, \(\begin{vmatrix} -1+3 & 2-1 & 5-5 \\ -3 & 1 & 5 \\ -1 & 2 & 5 \end{vmatrix}=\begin{vmatrix} 2 & 1 & 0 \\ -3 & 1 & 5 \\ -1 & 2 & 5 \end{vmatrix}\)
= 2 (5 -10) - 1 (-15 + 5) + 0
= -10 + 10 = 0
Thus, given lines are coplanar.
Now, equation of plane containing these lines is:
\(\begin{vmatrix} x+3 & y-1 & x-5 \\ -3 & 1 & 5 \\ -1 & 2 & 5 \end{vmatrix}=0\)
\(\Rightarrow\) (x + 3) (5 -10) - (y -1) (-15 + 5) + (z - 5) (- 6 + 1) = 0
\(\Rightarrow\) - 5 (x + 3) + 10(y-l) -5(z -5) = 0
\(\Rightarrow\) - 5x + 10y - 5z = 0
\(\Rightarrow\)  x - 2y + z = 0.

As on the y-axis, x-coordinate and z-coordinate are zeroes

As \({ cos }^{ 2 }\alpha +{ cos }^{ 2 }\beta +{ cos }^{ 2 }\gamma =1\)
\(\Rightarrow \frac { 1+cos2\alpha }{ 2 } +\frac { 1+cos2\beta }{ 2 } +\frac { 1+cos2\gamma }{ 2 } =1\)
= cos 2α + cos 2β + cos 2γ = 1

As direction cosines of a line whose direction ratio are 2,3, -6 are
\(\frac { -2 }{ 7 } ,\frac { 3 }{ 7 } ,\frac { 6 }{ 7 } \)
As angle with the y-axis is obtuse,
∴ cos β < 0,
Therefore direction ratios are \(\frac { -2 }{ 7 } ,\frac { -3 }{ 7 } ,\frac { 6 }{ 7 } \)

As let P divides the join of (3, 5, -1) and (6, 3, -2) in the ratio k : 1
\(\therefore \frac { 3k+5 }{ k+1 } =2\)
\(\Rightarrow 3k+5=2k+2\Rightarrow k=-3\)
∴ x-coordinate is
\(\frac { 6k+3 }{ k+1 } =\frac { -18+3 }{ -3+1 } =\frac { 15 }{ 2 } \)

As distance of point (2, 5, 7) from the x-axis is
\(\sqrt { { 5 }^{ 2 }+{ 7 }^{ 2 } } =\sqrt { 25+49 } =\sqrt { 74 } \)