It is known that the equation of the line passing through the points, \( \left(x_{1}, y_{1}, z_{1}\right) \text { and }\left(x_{2},\right.\left.y_{2}, z_{2}\right), \text { is } \frac{x-x_{1}}{x_{2}-x_{1}}=\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{z-z_{1}}{z_{2}-z_{1}}\)
The line passing through the points, (5,1,6) and (3,4,1), is given by,
\(\frac{x-5}{3-5}=\frac{y-1}{4-1}=\frac{z-6}{1-6} \)
\(\Rightarrow \frac{x-5}{-2}=\frac{y-1}{3}=\frac{z-6}{-5}=k(\text { say }) \)
\(\Rightarrow x=5-2 k, y=3 k+1, z=6-5 k \)
Any point on the line is of the form \((5-2 k, 3 k+1,6-5 k) \)
Since the line passes through ZX-plane,
\(3 k+1=0 \)
\(\Rightarrow k=-\frac{1}{3} \)
\(\Rightarrow 5-2 k=5-2\left(-\frac{1}{3}\right)=\frac{17}{3} \)
\(6-5 \mathrm{k}=6-5\left(-\frac{1}{3}\right)=\frac{23}{3} \)
Therefore, the required point is \( \left(\frac{17}{3}, 0, \frac{23}{3}\right) . \)

Clearly, the plane passes through the point (0, 3, 0)
The direction cosines of the normal to the plane are <0, 1, 0>
The equation of the plane is:
0.(x-0)+1.(y-3) + 0.(z-0) = 0
\(\Rightarrow\) y - 3 = 0
\(\Rightarrow\)y = 3.

If A(1, -1, 2) and B(3, 4, -2) be the given points, then the direction-ratios of AB are:
(3-1, 4-(-1),-2-2) i.e. (2, 5, -4)
Again if C(0, 3, 2) and D(3, 5, 6) be the given points, then the direction ratios of CD are:
(3-0, 5-3, 6-2) i.e.(3, 2, 4)
Now AB is perp. to CD are:
(2) (3)+(5)(2)+(-4)(4) = 0 [ a₁a₂ + b₁b₂ + c₁c₂ = 0]
i.e. if 6 +10-16 = 0, which is true.
Hence, AB is perpendicular to CD.

The angle between two planes is the angle between their normals. From the equation of the planes, the normal vectors are
\(\overrightarrow{\mathrm{N}}_{1}=2 \hat{i}+\hat{j}-2 \hat{k} \text { and } \overrightarrow{\mathrm{N}}_{2}=3 \hat{i}-6 \hat{j}-2 \hat{k}\)
\(\text {Therefore } \cos \theta=\left|\frac{\overrightarrow{\mathrm{N}}_{1} \cdot \overrightarrow{\mathrm{N}}_{2}}{\left|\overrightarrow{\mathrm{N}}_{1}\right|\left|\overrightarrow{\mathrm{N}}_{2}\right|}\right|=\left|\frac{(2 \hat{i}+\hat{j}-2 \hat{k}) \cdot(3 \hat{i}-6 \hat{j}-2 \hat{k})}{\sqrt{4+1+4} \sqrt{9+36+4}}\right|=\left(\frac{4}{21}\right)\)
\(\text {Hence }\theta=\cos ^{-1}\left(\frac{4}{21}\right) \)

Direction cosines are: \(\left< \cos { 90° } ,\cos { 60° } ,\cos { 30° } \right> (i.e.,<0,\frac { 1 }{ 2 } ,\frac { \sqrt { 3 } }{ 2 } >)\)

Equation of line PR is \(\frac { x-7 }{ 2 } =\frac { y-14 }{ 4 } =\frac { z-5 }{ -1 } =\lambda \)
General point on line is R \((2\lambda +7,4\lambda +14,-\lambda +5)\)
If this point lies in plane 2x + 4y - z =  2,
then \(2(2\lambda +7)+4(4\lambda +14)-(-\lambda +5)=2\Rightarrow 21\lambda =-63\Rightarrow \lambda =-3\)
Substituting in (i), Foot of perpendicular is R (1,2,8).
\(PR=\sqrt { { (7-1) }^{ 2 }+{ (14-2) }^{ 2 }+{ (5-8) }^{ 2 } } =\sqrt { 36+144+9 } =\sqrt { 189 } \) units
From (i) and (iii), we get 
\(\left( \frac { 7+\alpha }{ 2 } ,\frac { 14+\beta }{ 2 } ,\frac { 5\gamma }{ 2 } \right) =(1,2,8)\)
\(\Rightarrow \frac { 7+\alpha }{ 2 } =1,\frac { 14+\beta }{ 2 } =2,\frac { 5+\gamma }{ 2 } =8\Rightarrow \alpha =-5,\beta =-10,\gamma =11.\)
Image of point P (7,14,5) in plane 2x + 4y - z= 2 is Q (-5, -10, 11).

The given plane is 2x - y + z + 1 = 0 ...(1}
Let P (3, 2, 1) be the given point.

Let M be the foot of perpendicular from P on plane (1).
Let P' be the image of P in the plane (1).
The equations of PM are 
\(\frac { x-3 }{ 2 } =\frac { y-2 }{ -1 } =\frac { z-1 }{ 1 } ...(2)\)
Any point on (2) is (3 + 2k, 2 - k, 1 + k) ...(3)
This point is M if it lies on (1)
if 2 (3 + 2k) - (2 - k) + (1 + k) + 1 = 0
if  6k = -6 if k = -1.
Putting in (3), the point M is :
(3 + 2 (- 1), 2 - (- 1), 1 - 1) i.e. (1, 3, 0). ,
Hence, the co-ordinates of M, the foot of perpendicular are (1, 3, 0).
And perpendicular distance = IPMI
\(\sqrt { { \left( 3-1 \right) }^{ 2 }+{ \left( 2-3 \right) }^{ 2 }+{ \left( 1-0 \right) }^{ 2 } } \)
\(=\sqrt { 4+1+1 } =\sqrt { 6 } units\)
If P' \(\left( \alpha ,\beta ,\gamma \right) \) is then:
\(\sqrt { { \left( 3-1 \right) }^{ 2 }+{ \left( 2-3 \right) }^{ 2 }+{ \left( 1-0 \right) }^{ 2 } } \)
\(=\sqrt { 4+1+1 } =\sqrt { 6 } units\)
Hence, the required image is P'(-1, 4, -1)

\(\sin { \theta } =\frac { 2\times 10+3\times 2+6\times (-11) }{ \sqrt { 4+9+36 } \sqrt { 100+4+121 } } =\frac { 20+6-66 }{ 7\times 15 } =\frac { -8 }{ 21 }\)
\( \Rightarrow \theta ={ sin }^{ -1 }\left( \frac { -8 }{ 21 } \right) \)units

We know that the equation of the plane having intercepts a, band c on the three
co-ordinate axes is
\({{x}\over{a}}+{{y}\over{b}}+{{y}\over{c}}=1\)
Here, the co-ordinates of A, Band C are (a, 0, 0),  (0, b, 0) and (0, 0, c) respectively.
The centroid of \(\triangle \)ABC is \(\left({{a}\over{3}},{{b}\over{3}},{{c}\over{3}}\right).\)
Equating \(\left( {{a}\over{3}},{{b}\over{3}},{{c}\over{3}} \right)\) to \((\alpha,\beta,\gamma),\) we get a = \(3\alpha,\) b = \(3\beta\) and c = \(3\gamma\)
Thus, the equation of the plane is
\({{x}\over{3\alpha}}+{{y}\over{3\beta}}+{{z}\over{3\gamma}}=1\)
or \({{x}\over{\alpha}}+{{y}\over{\beta}}+{{z}\over{\gamma}}=3\)

Normal vector of the plane is
\(\overset { \rightarrow }{ x } =3\check { i } -2\check { j } +6\check { k } \)
\(\left| \overset { \rightarrow }{ x } \right| =\sqrt { 3^{ 2 }+2^{ 2 }+6^{ 2 } } \)
\( =\sqrt { 9+4+36 } =\sqrt { 49 } =7\)
\(\overset { \rightarrow }{ x } =\frac { \overset { \rightarrow }{ x } }{ \left| \overset { \rightarrow }{ x } \right| } =\frac { 3\check { i } -2\check { j } +6\check { k } }{ 7 } \)
The required equation of plane 
\(\Rightarrow \overset { \rightarrow }{ r } \frac { (3\check { i } +2\check { j } +6\check { k } ) }{ 7 } =5\)
\(\Rightarrow \check { r } .(3\check { i } +2\check { j } +6\check { k } )=35\)

The length of the perpendicular drawn from the origin to the plane 2x - 3y + 6z + 21 = 0
\(\frac { x }{ 5/2 } +\frac { y }{ 5 } +\frac { z }{ -5 } \)
\(=\frac { 21 }{ 7 } =3units\)

\(\frac { x-2 }{ 2 } =\frac { 2y-5 }{ -3 } ,z=-1\)
\(\Rightarrow \frac { x-2 }{ 2 } =\frac { 2y-5 }{ -3 } =\lambda ,z=-1+0\lambda \)
\(\frac { x-2 }{ 2 } =\frac { 2y-5 }{ -3 } ,z=-1\)
\(\Rightarrow \frac { x-2 }{ 2 } =\frac { y-5/2 }{ -3/2 } =\lambda ,z=-1+0\lambda \)
\(\Rightarrow \frac { x-2 }{ 2 } =\frac { y-5/2 }{ -3/2 } =\lambda ,\frac { z+1 }{ 0 } =\lambda \)
ratios are (2, - 3/2, 0) and the point on the given line is(2, 5/2, -1).

\( \left[ \begin{matrix} { l }_{ 4 } & { m }_{ 1 } & { n }_{ 1 } \\ { l }_{ 2 } & { m }_{ 2 } & { n }_{ 2 } \\ { l }_{ 3 } & { m }_{ 3 } & { n }_{ 3 } \end{matrix} \right] \left[ \begin{matrix} { l }_{ 1 } & { l }_{ 2 } & { l }_{ 3 } \\ { m }_{ 1 } & { m }_{ 2 } & { n }_{ 2 } \\ { n }_{ 1 } & { n }_{ 2 } & { n }_{ 3 } \end{matrix} \right] \)
\(\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix} \right] ={ l }_{ 2 }\)
because
\({ l }_{ 1 }^{ 2 }+{ m }_{ 1 }^{ 2 }+{ n }_{ 1 }^{ 2 }=1\) for each i = 1, 2, 3
\(l_{ i }l_{ 1 }+m_{ 1 }m_{ j }+n_{ i }n_{ j }=0(i=1)\quad for\quad each\quad i,j=1,2,3\)

\(\alpha =60^{ \circ },\beta =?,\gamma =45^{ \circ }\)
Let \(\alpha \) makes with x-axis, y makes with z-axis and \(\beta \) makes with y-axis
\(\alpha =cos\quad 60^{ \circ },m=cos\quad \beta ,y=cos45^{ \circ }\)
\(\alpha ^{ 2 }+\beta ^{ 2 }+\gamma ^{ 2 }=1\)
\(\Rightarrow cos^{ 2 }60^{ \circ }+cos^{ 2 }\beta +cos^{ 2 }45^{ \circ }\)
\(\Rightarrow \frac { 1 }{ 4 } +cos^{ 2 }\beta +\frac { 1 }{ 2 } =1\)
\(\Rightarrow cos^{ 2 }\beta =1-\frac { 1 }{ 4 } -\frac { 1 }{ 2 } \)
\(=\frac { 4-1-2 }{ 4 } \)
\(=\frac { 1 }{ 4 } \)
\(\Rightarrow cos \beta =\pm \frac { 1 }{ 2 } \)
\(\beta =\frac { \pi }{ 3 } or\ \frac { 2\pi }{ 3 } \)
But angle between two lines in the interval \(\left( 0,\frac { \pi }{ 2 } \right) \)
Hence required angles is \(\frac { \pi }{ 3 } \)

\(\theta=\sin ^{-1}\left(\frac{4}{\sqrt{42}}\right)\)

\(\frac { x+2 }{ 1 } =\frac { 2y-1 }{ 3 } =\frac { 3-z }{ 5 } .\)

Let equation of plane through (1, 2, -4) be a(x-1) + b(y-2) + c(z+4) = 0.
The plane is parallel to the given lines
\(\therefore\)  2a + 3b + 6c = 0; a + b - c = 0
Solving: \(\frac { a }{ -9 } =\frac { b }{ 8 } =\frac { c }{ -1 } =k\left( say \right) \)
\(\therefore\) a = -9k, b = 8k, c = -k
From (i), -9k(x-1) + 8k(y-2) - k(z+4) = 0
\(\therefore\) Equation of plane in cartesian form is 9x - 8y + z+11 = 0
Vector form of plane is: \(\Rightarrow \vec { r } .\left( 9\hat { i } -8\hat { j } +\hat { k } \right) =-11\)
Distance of (9, -8, -10) from the plane = \(\left| \frac { 9.9-8\left( -8 \right) +1\left( -10 \right) +11 }{ \sqrt { 81+64+1 } } \right| =\sqrt { 146 } \)