Three Dimensional Geometry

Vector perpendicular to face $A B C$.

$ \begin{aligned} & =A B \times A C=\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & 1 & 1 \\ 2 & -1 & 2 \end{array}\right| \\ & =\hat{\mathbf{i}}(2+1)-\hat{\mathbf{j}}(2-2)+\hat{\mathbf{k}}(-1-2) \\ & =3 \hat{\mathbf{i}}-0 \hat{\mathbf{j}}-3 \hat{\mathbf{k}} \end{aligned} $

Here, $A(1,2,1), B(2,3,2), C(3,1,3)$ and $D(2,1,3)$

$ \begin{aligned} & A B=O B-O A=(1,1,1) \\ & A C=(2,-1,2) \\ & A D=(1,-1,2) \text { and vector perpendicular to face } A B D \end{aligned} $

$ \begin{aligned} A B \times A D & =\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & 1 & 1 \\ 1 & -1 & 2 \end{array}\right| \\ & =\hat{\mathbf{i}}(2+1)-\hat{\mathbf{j}}(2-1)+\hat{\mathbf{k}}(-1-1)=3 \hat{\mathbf{i}}-\hat{\mathbf{j}}-2 \hat{\mathbf{k}} \end{aligned} $

Since, the angle between the faces $=$ Angle between their normals

Thus, $\cos \theta=\frac{9+0+6}{\sqrt{9+9} \sqrt{9+1+4}}$

$ \begin{aligned} & \cos \theta=\frac{15}{\sqrt{18} \sqrt{14}} \\ & \cos \theta=\frac{15}{\sqrt{9 \times 2} \sqrt{2 \times 7}} \\ & \cos \theta=\frac{15}{3 \sqrt{2} \sqrt{14}}=\frac{5}{\sqrt{2} \times \sqrt{2} \times \sqrt{7}} \\ & \cos \theta=\frac{5}{2 \sqrt{7}} \end{aligned} $

The tetrahedron is defined by the vertices $ A(3,2,4) $, $ B(x_1, y_1, 0) $, $ C(x_2, y_2, 0) $, and $ D(x_3, y_3, 0) $. The triangle $ \triangle BCD $ is formed by the lines $ y = x $, $ x + y = 6 $, and $ y = 1 $.

To find the centroid of the tetrahedron, we use the formula for the centroid of a tetrahedron with vertices $(x_1, y_1, z_1)$, $(x_2, y_2, z_2)$, $(x_3, y_3, z_3)$, and $(x_4, y_4, z_4)$:

$ \left(\frac{x_1+x_2+x_3+x_4}{4}, \frac{y_1+y_2+y_3+y_4}{4}, \frac{z_1+z_2+z_3+z_4}{4}\right) $

To determine the coordinates of the vertices $ B $, $ C $, and $ D $:

The intersection of the lines $ y = x $ and $ x + y = 6 $ gives the point $ (3, 3) $.

On the line $ y = x $, a point such as $ (5, 1) $ satisfies both conditions line $ x + y = 6 $.

A point on $ y = 1 $, such as $ (1, 1) $, also satisfies the line equation.

Thus, the coordinates of the vertices are:

$(x_2, y_2, z_2) = (3, 3, 0)$

$(x_3, y_3, z_3) = (5, 1, 0)$

$(x_4, y_4, z_4) = (1, 1, 0)$

Substituting these into the centroid formula gives:

$ \left(\frac{3 + 3 + 5 + 1}{4}, \frac{2 + 3 + 1 + 1}{4}, \frac{4 + 0 + 0 + 0}{4}\right) = \left(\frac{12}{4}, \frac{7}{4}, \frac{4}{4}\right) = \left(3, \frac{7}{4}, 1\right) $

Therefore, the centroid of the tetrahedron is $\left(3, \frac{7}{4}, 1\right)$.

Given, If $P(2, \beta, \alpha)$ lies on the plane $x+2 y-z-2=0$ and $Q(\alpha,-1, \beta)$ lies on the plane $2 x-y+3 z+6=0$ direction cosine of line $P Q=$ ?

Now, substitute the coordinates of $P$ into the plane equation $(x+2 y-z-2=0)$

$ 2+2 \beta-\alpha-2=0 \Rightarrow \alpha=2 \beta $

So, coordinate of $P$ are $(2, \beta, 2 \beta)$

Now, substitute the coordinates of $Q(\alpha,-1, \beta)$ into the plane equation

$ \begin{array}{r} (2 x-y+3 z+6=0) \\ 2 \alpha+1+3 \beta+6=0 \\ 2 \alpha+3 \beta+7=0 \end{array} $

Coordinate of $Q$ are $(\alpha,-1, \beta)$ direction ratios of the line $P Q$ are given by the differences in the coordinates of $Q$ and $P$.

$ P Q=(\alpha-2,-1-\beta, \beta-2 \beta) $

Here, $\alpha=-2$ and $\beta=-1$

From Eqs. (i) and (ii), we get

Magnitude of DR's

$ =\sqrt{(\alpha-2)^2+(-1-\beta)^2+(\beta-2 \beta)^2} $

Direction cosine of are $(l, m, n)$

$ \begin{aligned} & l=\frac{\alpha-2}{\sqrt{(\alpha-2)^2+(-1-\beta)^2+(\beta-2 \beta)^2}} \\ & m=\frac{-1-\beta}{\sqrt{(\alpha-2)^2+(-1-\beta)^2+(\beta-2 \beta)^2}} \\ & n=\frac{\beta-2 \beta}{\sqrt{(\alpha-2)^2+(-1-\beta)^2+(\beta-2 \beta)^2}} \\ & \text { So, } l=\frac{-2-2}{\sqrt{(-2-2)^2+(-1+1)^2+(1)^2}} \\ & \quad=\frac{-4}{\sqrt{17}} \\ & m=\frac{-1+1}{\sqrt{17}}=0 \\ & n=\frac{-1+2}{\sqrt{17}}=\frac{1}{\sqrt{17}}:\left(\frac{-4}{\sqrt{17}}, 0, \frac{1}{\sqrt{17}}\right) \end{aligned} $

To find the foot of the perpendicular from the point $ (1, 2, 1) $ to the plane $ \pi $, which passes through the point $(-2, 1, -1)$ and is parallel to the plane $2x - y + 2z = 0$, we can follow these steps:

Step 1: Determine the Equation of the Plane $\pi$

The plane $\pi$ has the same normal vector as the plane $2x - y + 2z = 0$, which is $2\hat{\mathbf{i}} - \hat{\mathbf{j}} + 2\hat{\mathbf{k}}$.

So, the general equation of the plane $\pi$ will be:

$ 2x - y + 2z = d $

Given that the plane passes through the point $(-2, 1, -1)$, we substitute these coordinates into the equation:

$ 2(-2) - 1 + 2(-1) = d $

$ -4 - 1 - 2 = d $

$ d = -7 $

Thus, the equation of the plane $\pi$ is:

$ 2x - y + 2z = -7 $

Step 2: Find the Foot of the Perpendicular

Let $ P(a, b, c) $ be the foot of the perpendicular from the point $ Q(1, 2, 1) $ to the plane. The vector from $ Q $ to $ P $, $ \overrightarrow{QP} $, must be parallel to the normal vector of the plane, which is given by:

$ 2\hat{\mathbf{i}} - \hat{\mathbf{j}} + 2\hat{\mathbf{k}} $

Thus, we can express:

$ \overrightarrow{QP} = (a - 1)\hat{\mathbf{i}} + (b - 2)\hat{\mathbf{j}} + (c - 1)\hat{\mathbf{k}} $

Equating components, we get the parallel condition:

$ \frac{a - 1}{2} = \frac{b - 2}{-1} = \frac{c - 1}{2} = \lambda $

From here, solve for $ a $, $ b $, and $ c $:

$ a = 2\lambda + 1 $

$ b = 2 - \lambda $

$ c = 2\lambda + 1 $

Step 3: Use the Plane Equation

Since point $ P(a, b, c) $ lies on the plane:

$ 2(2\lambda + 1) - (2 - \lambda) + 2(2\lambda + 1) = -7 $

$ 4\lambda + 2 - 2 + \lambda + 4\lambda + 2 = -7 $

$ 9\lambda + 2 = -7 $

Solving for $\lambda$:

$ 9\lambda = -9 $

$ \lambda = -1 $

Step 4: Substitute to Find Coordinates

Substituting $\lambda = -1$ back into the equations for $a$, $b$, and $c$:

$ a = 2(-1) + 1 = -1 $

$ b = 2 - (-1) = 3 $

$ c = 2(-1) + 1 = -1 $

Thus, the coordinates of the foot of the perpendicular are:

$ P(-1, 3, -1) $

Given the direction ratios $(2, 5, 1)$ of a line and the equation of the plane $8x + 2y - z = 14$, we aim to find the angle between the line and the plane.

To determine the angle $\theta$ between the line and the plane, we use the formula for the sine of the angle:

$ \sin \theta = \frac{a_1b_1 + a_2b_2 + a_3b_3}{\sqrt{a_1^2 + a_2^2 + a_3^2} \cdot \sqrt{b_1^2 + b_2^2 + b_3^2}} $

Here, the direction ratios of the line are $(a_1, a_2, a_3) = (2, 5, 1)$ and the normal to the plane, derived from the coefficients of $x$, $y$, and $z$, is $(b_1, b_2, b_3) = (8, 2, -1)$.

Substituting these into the formula, we calculate:

$ \sin \theta = \frac{2 \times 8 + 5 \times 2 + 1 \times (-1)}{\sqrt{2^2 + 5^2 + 1^2} \cdot \sqrt{8^2 + 2^2 + (-1)^2}} $

Simplifying, we get:

$ = \frac{16 + 10 - 1}{\sqrt{4 + 25 + 1} \cdot \sqrt{64 + 4 + 1}} $

$ = \frac{25}{\sqrt{30} \cdot \sqrt{69}} $

This simplifies further to:

$ = \frac{25}{3 \sqrt{230}} = \frac{25}{\sqrt{2070}} $

Therefore, the angle $\theta$ is given by:

$ \theta = \sin^{-1}\left(\frac{25}{3 \sqrt{230}}\right) = \sin^{-1}\left(\frac{25}{\sqrt{2070}}\right) $

To find the direction cosines of the line of intersection of the planes, we start with the given equations:

$ x + 2y + z - 4 = 0 $

$ 2x - y + z - 3 = 0 $

The direction ratios for these planes are derived from their coefficients, so for equation (i), the direction ratios are $ (1, 2, 1) $ and for equation (ii), they are $ (2, -1, 1) $.

To find the direction ratios of the line of intersection, we calculate the cross product of these direction ratios using:

$ \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & 2 & 1 \\ 2 & -1 & 1 \end{vmatrix} $

Calculating the cross product:

$ \begin{aligned} &= (2 \cdot 1 + 1 \cdot 1) \hat{\mathbf{i}} - (1 \cdot 1 - 2 \cdot 1) \hat{\mathbf{j}} + (1 \cdot (-1) - 2 \cdot 2) \hat{\mathbf{k}} \\ &= (2 + 1) \hat{\mathbf{i}} - (1 - 2) \hat{\mathbf{j}} + (-1 - 4) \hat{\mathbf{k}} \\ &= 3 \hat{\mathbf{i}} + \hat{\mathbf{j}} - 5 \hat{\mathbf{k}} \end{aligned} $

Thus, the direction ratios are $ \langle 3, 1, -5 \rangle $.

To convert these into direction cosines, we normalize the vector:

$ \left\langle \frac{3}{\sqrt{9+1+25}}, \frac{1}{\sqrt{9+1+25}}, \frac{-5}{\sqrt{9+1+25}} \right\rangle = \left\langle \frac{3}{\sqrt{35}}, \frac{1}{\sqrt{35}}, \frac{-5}{\sqrt{35}} \right\rangle $

These are the direction cosines of the line of intersection.

Given the direction ratios for lines $ L_1 $ and $ L_2 $ as $(3, 1, -5)$ and $(2, 3, -1)$ respectively, we need to find the equation of the plane that contains both lines $ L_1 $ and $ L_2 $.

To derive the equation of the plane, we use the determinant method with the direction ratios of the lines as follows:

$ \left|\begin{array}{ccc} x & y & z \\ 3 & 1 & -5 \\ 2 & 3 & -1 \\ \end{array}\right| = 0 $

This expands to:

$ x((-1 \times 3) - (15 \times 1)) - y((-3 \times 1) - (10 \times 3)) + z((9 \times 3) - (2 \times (-1))) $

After simplifying, we obtain:

$ x(-1 + 15) - y(-3 + 10) + z(9 - 2) = 0 $

Which further simplifies to:

$ 14x - 7y + 7z = 0 $

Dividing through by 7, we get the simplified equation of the plane:

$ 2x - y + z = 0 $

To find the position vector of point $ R $, where a line passing through points $ A $ and $ B $ intersects the plane defined by points $ O $, $ C $, and $ D $, we start by calculating the vector equations and the intersection.

Position vectors of the points are:

$ A = \hat{\mathbf{i}} + 2 \hat{\mathbf{j}} + \hat{\mathbf{k}} $

$ B = -2 \hat{\mathbf{i}} + 3 \hat{\mathbf{k}} $

$ C = 2 \hat{\mathbf{i}} + \hat{\mathbf{j}} $

$ D = 4 \hat{\mathbf{k}} $

Step 1: Find the vector $\mathbf{AB}$:

$ \mathbf{AB} = B - A = (-2 \hat{\mathbf{i}} + 3 \hat{\mathbf{k}}) - (\hat{\mathbf{i}} + 2 \hat{\mathbf{j}} + \hat{\mathbf{k}}) = -3 \hat{\mathbf{i}} - 2 \hat{\mathbf{j}} + 2 \hat{\mathbf{k}} $

Step 2: Parametric equation of the line through $ A $ and $ B $:

$ \mathbf{r} = \mathbf{A} + t \mathbf{AB} = (\hat{\mathbf{i}} + 2 \hat{\mathbf{j}} + \hat{\mathbf{k}}) + t (-3 \hat{\mathbf{i}} - 2 \hat{\mathbf{j}} + 2 \hat{\mathbf{k}}) $

Simplifying, we get:

$ \mathbf{r} = (1 - 3t) \hat{\mathbf{i}} + (2 - 2t) \hat{\mathbf{j}} + (1 + 2t) \hat{\mathbf{k}} $

Step 3: Equation for the plane through $ O $, $ C $, and $ D $:

$\mathbf{OC} = C - O = 2 \hat{\mathbf{i}} + \hat{\mathbf{j}}$

$\mathbf{OD} = D - O = 4 \hat{\mathbf{k}}$

Calculate the normal vector $\mathbf{n}$:

$ \mathbf{n} = \mathbf{OC} \times \mathbf{OD} = \left| \begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 2 & 1 & 0 \\ 0 & 0 & 4 \end{array} \right| = 4 \hat{\mathbf{i}} - 8 \hat{\mathbf{j}} $

Step 4: Plane equation based on $\mathbf{n}$:

The plane equation is:

$ 4x - 8y = 0 $

Step 5: Substitute the parametric line equation into the plane equation:

$ 4(1 - 3t) - 8(2 - 2t) = 0 $

Simplify the equation:

$ 4 - 12t - 16 + 16t = 0 \\ 4t - 12 = 0 \\ t = 3 $

Step 6: Calculate the coordinates of $ R $:

Substitute $ t = 3 $ into $\mathbf{r} = (1 - 3t) \hat{\mathbf{i}} + (2 - 2t) \hat{\mathbf{j}} + (1 + 2t) \hat{\mathbf{k}}$:

$ x = 1 - 3(3) = -8 $

$ y = 2 - 2(3) = -4 $

$ z = 1 + 2(3) = 7 $

Thus, the position vector of $ R $ is:

$ -8 \hat{\mathbf{i}} - 4 \hat{\mathbf{j}} + 7 \hat{\mathbf{k}} $

The distance of the point $ O(\mathbf{O}) $ from the plane, given by the equation $\mathbf{r} \cdot (\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}) = 5$, is measured along the direction of the vector $2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-6 \hat{\mathbf{k}}$.

First, we identify the plane's equation:

$ x + y + z = 5 $

The line parallel to the vector $2 \hat{\mathbf{i}} + 3 \hat{\mathbf{j}} - 6 \hat{\mathbf{k}}$ can be represented parametrically as:

$ \frac{x}{2} = \frac{y}{3} = \frac{z}{-6} = K $

This gives us:

$ x = 2K, \quad y = 3K, \quad z = -6K $

For the point $(2K, 3K, -6K)$ to be on the plane, it must satisfy the plane equation:

$ 2K + 3K - 6K = 5 $

Solving this, we find:

$ -K = 5 \quad \Rightarrow \quad K = -5 $

Thus, the coordinates of the point on the line are:

$ (-10, -15, 30) $

Now, to find the distance of this point from the origin, we calculate the Euclidean distance:

$ \begin{aligned} \text{Distance} &= \sqrt{(-10)^2 + (-15)^2 + 30^2} \\ &= \sqrt{100 + 225 + 900} \\ &= \sqrt{1225} \\ &= 35 \\ \end{aligned} $

Therefore, the required distance is 35.

To find the point of intersection $ P(a, b, c) $ of the lines $ AB $ and $ CD $, let's first determine the equations of these lines.

Equation of Line $ AB $

For line $ AB $, given points $ A(1,0,2) $ and $ B(2,1,0) $, the direction ratios are:

$ (2-1, 1-0, 0-2) = (1, 1, -2) $

Thus, the equation of line $ AB $ is:

$ \frac{x-1}{1} = \frac{y-0}{1} = \frac{z-2}{-2} $

Equation of Line $ CD $

For line $ CD $, given points $ C(2,-5,3) $ and $ D(0,3,2) $, the direction ratios are:

$ (0-2, 3+5, 2-3) = (-2, 8, -1) $

Thus, the equation of line $ CD $ is:

$ \frac{x}{-2} = \frac{y-3}{8} = \frac{z-2}{-1} = K $

Therefore, in parametric form, the point $ P $ on line $ CD $ can be expressed as:

$ (x, y, z) = (2K, 3 - 8K, K + 2) $

Point $ P $ on Line $ AB $

Since point $ P $ also lies on line $ AB $, we equate the expressions:

$ 2K - 1 = 3 - 8K = \frac{K}{-2} $

Solving the first two equations:

$ 2K - 1 = 3 - 8K $

$ 10K = 4 $

$ K = \frac{2}{5} $

Substituting $ K = \frac{2}{5} $ into the parametric form:

$ x = 2K = 2 \times \frac{2}{5} = \frac{4}{5} $

$ y = 3 - 8K = 3 - 8 \times \frac{2}{5} = -\frac{1}{5} $

$ z = K + 2 = \frac{2}{5} + 2 = \frac{12}{5} $

Thus, the coordinates of $ P $ are:

$ \left(\frac{4}{5}, -\frac{1}{5}, \frac{12}{5}\right) $

Therefore, the sum $ a + b + c $ is:

$ a + b + c = \frac{4}{5} - \frac{1}{5} + \frac{12}{5} = 3 $

We start with the relations for the direction cosines of two lines:

$ l+m-n=0 \quad \text{and} \quad lm - 2mn + nl = 0 $

From $ l+m-n=0 $, we can express $ n $ as:

$ n = l + m $

Substituting this into the second equation, we have:

$ lm - 2m(l+m) + (l+m)l = 0 $

This simplifies to:

$ lm - 2lm - 2m^2 + l^2 + lm = 0 $

$ l^2 = 2m^2 $

Thus, $ l = \pm \sqrt{2}m $.

Now, we express $ n $ based on these values of $ l $:

$ n = \pm \sqrt{2}m + m = (\pm \sqrt{2} + 1)m $

We deduce two scenarios for the direction cosines:

$ l:m:n = \sqrt{2}:1:\sqrt{2}+1 $

$ l:m:n = -\sqrt{2}:1:-\sqrt{2}+1 $

The cosine of the acute angle $ \theta $ between the lines is given by:

$ \cos \theta = \frac{\sqrt{2} \times (-\sqrt{2}) + 1 \times 1 + (\sqrt{2} + 1)(-\sqrt{2} + 1)}{\sqrt{2+1+2+1+2\sqrt{2}} \sqrt{2+1+2+1-2\sqrt{2}}} $

This simplifies as:

$ = \frac{|-2 + 1 - 1|}{\sqrt{6 + 2\sqrt{2}} \sqrt{6 - 2\sqrt{2}}} = \frac{2}{\sqrt{36 - 8}} = \frac{1}{\sqrt{7}} $

Therefore, the cosine of the acute angle $ \theta $ is $\frac{1}{\sqrt{7}}$.

To determine the locus of point $ P $, consider a plane $\pi$ intersecting the coordinate axes at points $ A $, $ B $, and $ C $.

The equation of the plane is given by:

$ \frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1 $

where the plane intersects the $ X $-axis at $ A(a, 0, 0) $, $ Y $-axis at $ B(0, b, 0) $ and $ Z $-axis at $ C(0, 0, c) $.

The distance from a fixed point $(1, 1, 1)$ to this plane is 12 units. The formula for the distance $ d $ from a point $(x_1, y_1, z_1)$ to a plane $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$ is given by:

$ d = \frac{\left| \frac{x_1}{a} + \frac{y_1}{b} + \frac{z_1}{c} - 1 \right|}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}}} $

Substituting the point $(1, 1, 1)$ into the distance formula, we have:

$ \frac{\left| \frac{1}{a} + \frac{1}{b} + \frac{1}{c} - 1 \right|}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}}} = 12 $

Squaring both sides, we obtain:

$ \left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c} - 1\right)^2 = 144 \left(\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}\right) $

Thus, we identify that point $ P $, which is the intersection of planes through points $ A $, $ B $, $ C $ and parallel to the coordinate planes, adheres to the following equation:

$ \left(\frac{1}{x} + \frac{1}{y} + \frac{1}{z} - 1\right)^2 = 144 \left(\frac{1}{x^2} + \frac{1}{y^2} + \frac{1}{z^2}\right) $

Therefore, the locus of $ P $ satisfies this equation.

To find the shortest distance between two skew lines given by:

$\mathbf{r} = (-\hat{\mathbf{i}} - 2\hat{\mathbf{j}} - 3\hat{\mathbf{k}}) + t(3\hat{\mathbf{i}} - 2\hat{\mathbf{j}} - 2\hat{\mathbf{k}})$

$\mathbf{r} = (7\hat{\mathbf{i}} + 4\hat{\mathbf{k}}) + s(\hat{\mathbf{i}} - 2\hat{\mathbf{j}} + 2\hat{\mathbf{k}})$

follow these steps:

Define the Position Vectors and Direction Vectors:

$\mathbf{a}_1 = -\hat{\mathbf{i}} - 2\hat{\mathbf{j}} - 3\hat{\mathbf{k}}$

$\mathbf{b}_1 = 3\hat{\mathbf{i}} - 2\hat{\mathbf{j}} - 2\hat{\mathbf{k}}$

$\mathbf{a}_2 = 7\hat{\mathbf{i}} + 4\hat{\mathbf{k}}$

$\mathbf{b}_2 = \hat{\mathbf{i}} - 2\hat{\mathbf{j}} + 2\hat{\mathbf{k}}$

Calculate the Vector Connecting Points on the Lines:

$ \mathbf{a}_2 - \mathbf{a}_1 = (7 + 1)\hat{\mathbf{i}} + (0 + 2)\hat{\mathbf{j}} + (4 + 3)\hat{\mathbf{k}} = 8\hat{\mathbf{i}} + 2\hat{\mathbf{j}} + 7\hat{\mathbf{k}} $

Find the Cross Product of the Direction Vectors:

$ \mathbf{b}_1 \times \mathbf{b}_2 = \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 3 & -2 & -2 \\ 1 & -2 & 2 \end{vmatrix} $

$ = \hat{\mathbf{i}}((-2) \cdot 2 - (-2) \cdot (-2)) - \hat{\mathbf{j}}(3 \cdot 2 - (-2) \cdot 1) + \hat{\mathbf{k}}(3 \cdot (-2) - (-2) \cdot 1) $

$ = \hat{\mathbf{i}}(-4 - 4) - \hat{\mathbf{j}}(6 + 2) + \hat{\mathbf{k}}(-6 + 2) $

$ = -8\hat{\mathbf{i}} - 8\hat{\mathbf{j}} - 4\hat{\mathbf{k}} $

Calculate the Magnitude of the Cross Product:

$ \left|\mathbf{b}_1 \times \mathbf{b}_2\right| = \sqrt{(-8)^2 + (-8)^2 + (-4)^2} = \sqrt{64 + 64 + 16} = \sqrt{144} = 12 $

Find the Required Distance:

$ \text{Distance} = \frac{|(8\hat{\mathbf{i}} + 2\hat{\mathbf{j}} + 7\hat{\mathbf{k}}) \cdot (-8\hat{\mathbf{i}} - 8\hat{\mathbf{j}} - 4\hat{\mathbf{k}})|}{12} $

$ = \frac{|8(-8) + 2(-8) + 7(-4)|}{12} $

$ = \frac{|-64 - 16 - 28|}{12} $

$ = \frac{108}{12} = 9 $

Thus, the shortest distance between these skew lines is 9.

Let $A E$ bisects $\angle B A C$.

$ \begin{aligned} A B & =\sqrt{(2-1)^2+(0-2)^2+(1-0)^2}=\sqrt{6} \\ A C & =\sqrt{(-3-1)^2+(0-2)^2+(2-0)^2} \\ & =\sqrt{24}=2 \sqrt{6} \\ \frac{B E}{E C} & =\frac{A B}{A C}=\frac{\sqrt{6}}{2 \sqrt{6}}=\frac{1}{2} \end{aligned} $

$\therefore E$ divides $B C$ in ratio 1:2

By section formula,

Coordinate of $E$ is

$ \begin{aligned} & \left(\frac{-3+4}{1+2}, \frac{0+0}{1+2}, \frac{2+2}{1+2}\right) \\ & \Rightarrow \\ & \begin{aligned} & A B\left.=\sqrt{\left(1-\frac{1}{3}, 0, \frac{4}{3}\right)}\right)^2+(2-0)^2+\left(0-\frac{4}{3}\right)^2 \\ &=\sqrt{\frac{4}{9}+4+\frac{16}{9}} \\ & \quad=\sqrt{\frac{56}{9}}=\frac{2 \sqrt{14}}{3} \end{aligned} \end{aligned} $

Equation of line joining $(0,2,4)$ and $(3,0,1)$ is

$ \frac{x}{3}=\frac{y-2}{-2}=\frac{z-4}{-3}=k $

Let coordinate of foot of perpendicular be $M$ ( $3 k, 2-2 k, 4-3 k$ )

Direction ratio of $B C=\langle 3,-2,-3\rangle$

Direction ratio of

$ A M=<3 k+1,1-2 k, 4-3 k\rangle $

As $A M$ and $B C$ are perpendicular

$ \begin{aligned} & 3(3 k+1)-2(1-2 k)-3(4-3 k)=0 \\ & 9 k+3-2+4 k-12+9 k=0 \\ & 22 k-11=0 \\ & \quad k=\frac{1}{2} \end{aligned} $

$\therefore$ Coordinate of $M$ is $\left(\frac{3}{2}, 1, \frac{5}{2}\right)$

$ \begin{aligned} & \text { Hence, required length }=A M \\ & =\sqrt{\left(-1-\frac{3}{2}\right)^2+(1-1)^2+\left(0-\frac{5}{2}\right)^2} \\ & =\sqrt{\frac{25}{4}+\frac{25}{4}}=\sqrt{2} \times \frac{5}{2}=\frac{5}{\sqrt{2}} \end{aligned} $

To find the relationship between the line and the plane, first determine their respective equations.

Equation of the Line:

The line $L$ passes through the points $(1, 2, -3)$ and $(\beta, 3, 1)$, but let's assume the correct coordinates at the second point are $(3, 3, -1)$ based on common calculation practices for vector direction. This gives the direction vector as $(3 - 1, 3 - 2, -1 + 3) = (2, 1, 2)$.

The parametric equations of the line can be written as:

$ \frac{x - 1}{2} = \frac{y - 2}{1} = \frac{z + 3}{2} $

Equation of the Plane:

The plane $\pi$ passes through the points $(2, 1, -2)$, $(-2, -3, 6)$, and $(0, 2, -1)$. The normal vector of the plane can be computed using the cross product of two vectors formed by these points:

Vector 1: $(-2 - 2, -3 - 1, 6 + 2) = (-4, -4, 8)$

Vector 2: $(0 - 2, 2 - 1, -1 + 2) = (-2, 1, 1)$

The cross product (normal vector) is:

$ \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -4 & -4 & 8 \\ -2 & 1 & 1 \end{vmatrix} = \mathbf{i}((1)(1) - (1)(8)) - \mathbf{j}((-4)(1) - (8)(-2)) + \mathbf{k}((-4)(1) - (-4)(-2)) $

Solving, we find:

$ \mathbf{i}(-4) - \mathbf{j}(12) + \mathbf{k}(-12) = (-4, -12, -12) $

Thus, the plane equation is:

$ -4(x - 2) - 12(y - 1) - 12(z + 2) = 0 $

Expanding and simplifying, we get:

$ -4x - 12y - 12z + 8 + 12 + 24 = 0 $

$ -4x - 12y - 12z + 44 = 0 $

It simplifies to:

$ x + y + z = 5 $

Angle Between Line and Plane:

The direction vector of the line is $(2, 1, 2)$ and the normal vector of the plane is $(1, 1, 1)$.

The angle $\theta$ between the line and the plane is given by:

$ \sin \theta = \frac{|2 \cdot 1 + 1 \cdot 1 + 2 \cdot 1|}{\sqrt{2^2 + 1^2 + 2^2} \sqrt{1^2 + 1^2 + 1^2}} = \frac{|2 + 1 + 2|}{\sqrt{9} \cdot \sqrt{3}} = \frac{5}{3\sqrt{3}} $

Now, compute $\cos^2 \theta$:

$ 27 \cos^2 \theta = 27 (1 - \sin^2 \theta) $

Substitute $\sin^2 \theta$:

$ 1 - \sin^2 \theta = 1 - \left(\frac{5}{3\sqrt{3}}\right)^2 = 1 - \frac{25}{27} = \frac{2}{27} $

Therefore,

$ 27 \cos^2 \theta = 27 \times \frac{2}{27} = 2 $

In conclusion, the value is $\boxed{2}$.

To determine if the points with the given position vectors are collinear, we need to set up vectors $ \overrightarrow{AB} $ and $ \overrightarrow{AC} $ as follows:

Let point $ A $ have the position vector $ \alpha \hat{\mathbf{i}} + 10 \hat{\mathbf{j}} + 13 \hat{\mathbf{k}} $.

Let point $ B $ have the position vector $ 6 \hat{\mathbf{i}} + 11 \hat{\mathbf{j}} + 11 \hat{\mathbf{k}} $.

Let point $ C $ have the position vector $ \frac{9}{2} \hat{\mathbf{i}} + \beta \hat{\mathbf{j}} - 8 \hat{\mathbf{k}} $.

The vector $ \overrightarrow{AB} $ is calculated as:

$ \overrightarrow{AB} = (6 - \alpha) \hat{\mathbf{i}} + (11 - 10) \hat{\mathbf{j}} + (11 - 13) \hat{\mathbf{k}} $

So:

$ \overrightarrow{AB} = (6 - \alpha) \hat{\mathbf{i}} + \hat{\mathbf{j}} - 2 \hat{\mathbf{k}} $

The vector $ \overrightarrow{AC} $ is calculated as:

$ \overrightarrow{AC} = \left(\frac{9}{2} - \alpha\right) \hat{\mathbf{i}} + (\beta - 10) \hat{\mathbf{j}} + (-8 - 13) \hat{\mathbf{k}} $

So:

$ \overrightarrow{AC} = \left(\frac{9}{2} - \alpha\right) \hat{\mathbf{i}} + (\beta - 10) \hat{\mathbf{j}} - 21 \hat{\mathbf{k}} $

Since the points are collinear, vectors $ \overrightarrow{AB} $ and $ \overrightarrow{AC} $ are parallel, which implies:

$ \frac{6 - \alpha}{\frac{9}{2} - \alpha} = \frac{1}{\beta - 10} = \frac{-2}{-21} $

Solving these equations, we find:

$ \frac{1}{\beta - 10} = \frac{2}{21} \; \Rightarrow \; \beta - 10 = \frac{21}{2} $

$ \beta = 10 + \frac{21}{2} = \frac{41}{2} $

For the other ratio:

$ \frac{12 - 2\alpha}{9 - 2\alpha} = \frac{2}{21} $

$ 126 - 21\alpha = 9 - 2\alpha $

So:

$ 19\alpha = 117 $

$ \alpha = \frac{117}{19} = 6 $

Given:

$ (19\alpha - 6\beta)^2 = (117 - 123)^2 = (-6)^2 = 36 $

Thus, the final result is $(19\alpha - 6\beta)^2 = 36$.

The given equation is $ axy + by = cy $.

We can simplify this to:

$ ax + bz = c $

This equation represents a plane. The key characteristic of this plane is its orientation in space.

Since there is no specific term dealing with the $ y $-coordinate in the rearranged equation $ ax + bz = c $, this plane is parallel to the $ Y $-axis.

Thus, it is a plane that is perpendicular to the $ ZX $-plane or could coincide with the $ ZX $-plane itself.

In simpler terms, the $ ZX $-plane is defined where $ y = 0 $, and any plane parallel to it or identical will have no effect from changes in the $ y $-coordinate.

We know that $Y \mathbb{Z}$ plane divides $P Q$ in the ratio 2:3.

By section formula the intersecting point is

$ \left(\frac{3 a+6}{5}, \frac{12+2 \beta}{5}, \frac{21+16}{5}\right) $

On $Y$ plane, $x$-coordinate of any point is zera.

$ \Rightarrow 3 \alpha+6=0 \Rightarrow \alpha=-2 $

$\therefore$ Coordinate of $P$ is $(-24,7)$

Also, $2 X$ plane divides $P Q$ in ratio $4: 5$.

The intersecting point is

$ \left(\frac{12+5 a}{9}, \frac{4 \beta+20}{9}, \frac{32+35}{9}\right) $

On $2 x$ plane, $y$-coordinate of any point is 0 .

$ \Rightarrow \quad 4 \beta+20=0 \Rightarrow \beta=-5 $

$\therefore$ Coordinate of $Q$ is $(3,-5.8)$

$ \begin{aligned} P Q & =\sqrt{13+2)^2+(-5-4)^2+(8-7)^2} \\ & =\sqrt{25+81+1} \\ & =\sqrt{107} \end{aligned} $

Given, points $A(1,2,-1)$ and $B(-1,0,1)$ and $P$ divides the line segment externally in the ratio $1: 2$.

For external division, coordinates are

$\begin{aligned} & {\left[\frac{m_1 x_2-m_2 x_1}{m_1-m_2}, \frac{m_1 y_2-m_2 y_1}{m_1-m_2}, \frac{m_1 z_2-m_2 z_1}{m_1-m_2}\right]} \\ & \quad=\left[\frac{1(-1)-2(1)}{1-2}, \frac{1(0)-2(2)}{1-2}, \frac{1(1)-2(-1)}{1-2}\right] \\ & \quad=\left[\frac{-1-2}{-1}, \frac{-4}{-1}, \frac{3}{-1}\right] \\ & \quad=[3,4,-3] \end{aligned}$

$\begin{aligned} \text{So,}\quad PQ & =\sqrt{(3-1)^2+(4-3)^2+(-3+1)^2} \\ & =\sqrt{(2)^2+(1)^2+(-2)^2} \\ & =\sqrt{4+1+4}=3 \text { units } \end{aligned} $

Direction cosine of $a \hat{i}+b \hat{j}+c \hat{k}$ can be $\frac{a}{\sqrt{a^2+b^2+c^2}}, \frac{b}{\sqrt{a^2+b^2+c^2}}, \frac{c}{\sqrt{a^2+b^2+c^2}}$

According to question,

$\sqrt{a^2+b^2+c^2}=\sqrt{83}, b=5$

$\begin{array}{lr} \Rightarrow & \quad a^2+(5)^2+c^2=83 \\ \Rightarrow & a^2+c^2=83-25 \\ \Rightarrow & a^2+c^2=58 \quad \text{... (i)} \end{array}$

and $c-a=4\quad \text{[given]}$

$\begin{aligned} &\begin{aligned} & (c-a)^2+2 a c=a^2+c^2 \\ & \Rightarrow(4)^2+2 a c=58 \quad\text { [from Eq. (i), } c^2+a^2=58 \text { ] }\\ & \Rightarrow 2 a c=58-16 \\ & a c=\frac{42}{2}=21 \end{aligned}\\ \end{aligned}$

The equation of plane passes through the point $(1,0,1)$ is

$a(x-1)+b(y-0)+c(z-1)=0\quad \text{... (i)}$

If this plane is perpendicular to the planes $2 x+3 y-z=2 \text { and } x-y+2 z=1$

Then, $2 a+3 b-c=2$ and $a-b+2 c=1$

$\begin{aligned} & \Rightarrow \frac{a}{6-1}=\frac{b}{-1-4}=\frac{c}{2(-1)-3} \\ & \Rightarrow \frac{a}{5}=\frac{b}{-3}=\frac{c}{-5}=k \end{aligned}$

Let $a=5 k, b=-3 k, c=-5 k$

Putting the values of $a, b$ and $c$ in Eq. (i), we get the required equation of plane

$\begin{aligned} & 5 k(x-1)+(-3 k)(y)+(-5 k)(z-1)=0 \\ \Rightarrow & 5 x-5-3 y-5 z+5=0 \\ \Rightarrow & 5 x-3 y-5 z=0 \end{aligned}$

Now, plane passing through $(11,7,5)$ and parallel to the plane $\pi(5 x-3 y-5 z)$.

Equation of a plane parallel to

$5 x-3 y-5 z=k \quad \text{... (i)}$

and passing through $(11,7,5)$

$\begin{aligned} & & 5(11)-3(7)-5(5) & =k \\ \Rightarrow & & 55-21-25 & =k \\ \Rightarrow & & k & =9 \end{aligned}$

So, equation of plane

$\Rightarrow 5 x-3 y-5 z=9$

Comparing with $a x+b y-z+d=0$ (given)

So, $a=5, b=-3, c=5, d=9$

$\begin{aligned} & \frac{a}{b}+\frac{b}{d}=\frac{5}{-3}+\frac{-3}{9} \\ \Rightarrow & \frac{a}{b}+\frac{b}{d}=\frac{-15-3}{9}=\frac{-18}{9}=-2 \end{aligned}$

Let a, b and c are the position vectors of the vertices A, B are C, respectively.

Also $\mathbf{d}, \mathbf{e}$ and $\mathbf{f}$ be the position vectors of points $D, E$ and $F$, respectively

$\mathrm{d}=\frac{2 \mathrm{c}+3 \mathrm{~b}}{5}, \mathrm{e}=\frac{2 \mathrm{c}+\mathrm{a}}{3}, \mathrm{f}=\frac{3 \mathrm{~b}+\mathrm{a}}{4}$

Equation of line $B E$ is $\mathbf{r}=\mathbf{b}+k(\mathbf{e}-\mathbf{b})$,

$\begin{aligned} \mathbf{r} & =\mathbf{b}+k\left(\frac{2 \mathbf{c}+\mathbf{a}}{3}-\mathbf{b}\right) \\ \Rightarrow \quad \mathbf{r} & =\mathbf{b}+k\left[\frac{2 \mathbf{c}+\mathbf{a}-3 \mathbf{b}}{3}\right] \\ \Rightarrow \quad \mathbf{r} & =\frac{k}{3} \mathbf{a}+\mathbf{b}[1-k]+\frac{2 k}{3} \mathbf{c} \end{aligned}$

Equation of line $C F$ is

$\begin{aligned} \mathbf{r} & =\mathbf{c}+\lambda(\mathbf{f}-\mathbf{c}) \\ & =\mathbf{c}+\lambda\left(\frac{3 b+\mathbf{a}}{4}-\mathbf{c}\right)=\mathbf{c}+\lambda\left[\frac{3 b+\mathbf{a}-4 \mathbf{c}}{4}\right] \\ & =\frac{\lambda \mathbf{a}}{4}+\frac{3 \lambda \mathbf{b}}{4}+\mathbf{c}(1-\lambda) \end{aligned}$

$C F$ and $B E$ intersect at point $P$

$\frac{k \mathbf{a}}{3}+\mathbf{b}(1-k)+\frac{2 k}{3} \mathbf{c}=\frac{\lambda \mathbf{a}}{4}+\frac{3 \lambda \mathbf{b}}{4}+\mathbf{c}(1-\lambda)$

On comparing,

$\frac{k}{3}=\frac{\lambda}{4}$

$\Rightarrow \quad k=\frac{3 \lambda}{4}$

Also,

$\begin{array}{rlrl} & & 1-k & =\frac{3 \lambda}{4} \Rightarrow 1-k=k \\ \Rightarrow & & 2 k & =1 \Rightarrow k=\frac{1}{2} \\ \text{and} & & \frac{1}{2} & =\frac{3 \lambda}{4} \\ && \lambda & =\frac{2}{3} \end{array}$

Position vector of $P=\frac{\mathbf{a}}{6}+\frac{\mathbf{b}}{2}+\frac{\mathbf{c}}{3}=\frac{\mathbf{a}+3 \mathbf{b}+2 \mathbf{c}}{6}$

$\begin{aligned} & \mathbf{A P}=x_1 \mathbf{A B}+y_1 \mathbf{A C} \\ &\left(\frac{\mathbf{a}+3 \mathbf{b}+2 \mathbf{c}}{6}-\mathbf{a}\right)=x_1(\mathbf{b}-\mathbf{a})+y_1(\mathbf{c}-\mathbf{a}) \\ & \Rightarrow \frac{-5 \mathbf{a}+3 \mathbf{b}+2 \mathbf{c}}{6}=\mathbf{a}\left(-x_1-y_1\right)+x_1 \mathbf{b}+y_1 \mathbf{C} \\ & \therefore \quad-x_1-y_1=\frac{-5}{6} \\ & x_1+y_1=\frac{5}{6} \end{aligned}$

The equation of the plane passing through the point $A(-2,1,3)$ and perpendicular to the vector $3 \hat{i}+\hat{j}+5 \hat{k}$ is $3 x+y+5 z=d$.

Thus, $d=-6+1+15=10$

Then, $3 x+y+5 z-10=0$

Thus, $a=3, b=1, c=5, d=-10$

$\frac{a+b}{c+d}=\frac{3+1}{5-10}=-\frac{4}{5}$

Given, $Q(2,2,1)$ and $R(5,2,-2)$.

Thus, equation of line through $R$ and $Q$ is

$\begin{aligned} & \frac{x-2}{2-5}=\frac{y-2}{2-2}=\frac{z-1}{1+2} \\ \Rightarrow & \frac{x-2}{-3}=\frac{y-2}{0}=\frac{z-1}{3}=r \text { (say) } \end{aligned}$

Thus, $P$ be point on the line. So, $P=(-3 r+2), 2,(3 r+1)$

Since, $x$-coordinate is 4 . So, $-3 r+2=4$

$\Rightarrow \quad r=\frac{-2}{3}$

Thus, $y$-coordinate $=2$

and $z$-coordinate $=-1$

$\therefore y$-coordinate $=-2(z$ - coordinate of $P$)

Mid-point of $A$ and $B$ is $\left(\frac{p+3}{2}, \frac{-4+2}{2}, \frac{2-4}{2}\right) =\left(\frac{p+3}{2},-1,-1\right)$

Let $P$ be mid-point of $A$ and $B$.

DR'S of $C P=\frac{p+3}{2}-5,-1-q,-1-1$

$=\frac{p-7}{2},-1-q,-2$

$\begin{aligned} & \text { Here, } \frac{p-7}{2}=2,-1-q=3, c=-2 \\ & \Rightarrow \quad p=11, q=-4, c=-2 \\ & \Rightarrow \quad c(p+7 q)=-2[11-28]=-2(-17)=34 \end{aligned}$

Given, the plane normal to the vector

$\hat{n}=\hat{i}+2 \hat{j}+2 \hat{k}$

Let equation of plane

$x+2 y+2 z+d=0$

Distance of a plane from the origin is $1 / 3$.

Thus, distance $=\left|\frac{a x_1+b y_1+c z_1+d}{\sqrt{a^2+b^2+c^2}}\right|$

$\begin{aligned} & \Rightarrow \frac{1}{3}=\left|\frac{0+2 \cdot 0+2 \cdot 0+d}{\sqrt{1+2^2+2^2}}\right| \\ & \Rightarrow 1 / 3=d / 3 \Rightarrow d=1 \end{aligned}$

Thus, equation of the plane is $x+2 y+2 z+1=0$

Comparing given equation of the plane

$\begin{aligned} & x+p y+q z+r=0 \\ & \therefore \quad p=2, q=2, r=1 \\ & \therefore \quad \sqrt{p^2+q^2+r^2}=\sqrt{2^2+2^2+1^2} \\ & \Rightarrow \quad \sqrt{9}=3 \\ \end{aligned}$

$\mathbf{r}=2 \mathbf{b}+t(6 \mathbf{c}-\mathbf{a})$

and $\mathbf{r}=\mathbf{a}+s(\mathbf{b}-3 c)$

Put $t=-1$ and $s=24$ and check

If $t=-1 \Rightarrow \mathbf{r}=2 \mathrm{~b}-1(6 \mathrm{c}-\mathrm{a})=2 \mathrm{~b}+\mathrm{a}-6 \mathrm{c}$

If $s=2 \Rightarrow \mathbf{r}=\mathbf{a}+2(\mathbf{b}-3 \mathrm{c})=\mathbf{a}+2 \mathbf{b}-6 \mathbf{c}$

Therefore, we can say that $\mathbf{a}+2 b-6 c$ lies on both the lines, i.e. it is the intersecting point.

$\begin{aligned} x & =\frac{m x_2+n x_1}{m+n} \\ \Rightarrow \quad a & =\frac{5 m+n}{m+n} \\ \Rightarrow \quad a & =\frac{5\left(\frac{m}{n}\right)+1}{\frac{m}{n}+1} \quad \text{.... (i)} \end{aligned}$

$\begin{aligned} & y \text {-coordinate }: 8=\frac{2 m+4 n}{m+n} \\ & \Rightarrow \quad 8 m+8 n=2 m+4 n \\ & \Rightarrow \quad 6 m=-4 n \\ & \Rightarrow \quad 3 m=-2 n \quad \text{... (ii)}\\ \end{aligned}$

$z \text { - coordinate }:-2=\frac{10 m+6 n}{m+n}$

$\begin{aligned} \Rightarrow & -2 m-2 n =10 m+6 n \\ \Rightarrow & -8 n =12 m \Rightarrow-2 n=3 m \\ \Rightarrow & \frac{m}{n} =-\frac{2}{3}\quad \text{... (iii)} \end{aligned}$

From Eq. (iii), put $\frac{m}{n}=\frac{-2}{3}$ into Eq. (i),

$\begin{aligned} a & =\frac{5\left(-\frac{2}{3}\right)+1}{-\frac{2}{3}+1}=\frac{-10+3}{-2+3}=-7 \\ \therefore \quad & \frac{2 m}{n}-\frac{a}{3}=2\left(-\frac{2}{3}\right)-\left(-\frac{7}{3}\right) \\ & =\frac{-4}{3}+\frac{7}{3}=\frac{-4+7}{3}=\frac{3}{3}=1 \end{aligned}$

Given, points are $(4,3,-5)$ and $(-2,1,-8)$.

Direction ratios

$\begin{aligned} &(a, b, c)=(4-(-2), 3-1,-5-(-8))=(6,2,3) \\ & \therefore \quad P(a, 3 b, 2 c) \equiv(6,3 \times 2,2 \times 3) \equiv(6,6,6) \\ & \text { For } x+y-2 z=0 \\ & \Rightarrow \quad 6+6-2 \times 6=0 \\ & \Rightarrow \quad 12-12=0 \text { (true) } \end{aligned}$

Let plane's equation be

$\begin{aligned} \quad & \frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1 \quad \text{....(i)}\\ \Rightarrow \quad & a=\frac{5}{2} \end{aligned}$

and plane passes through $(1,1,1)$.

$\begin{array}{ll} \therefore & \frac{1}{5 / 2}+\frac{1}{b}+\frac{1}{c}=1 \\ \Rightarrow & \frac{1}{b}+\frac{1}{c}=1-\frac{2}{5} \\ \Rightarrow & \frac{1}{b}+\frac{1}{c}=\frac{3}{5} \quad \text{.... (ii)} \end{array}$

Perpendicular distance of plane from origin $=\frac{5}{7}$

$\begin{aligned} & \left|\frac{0+0+0-1}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}}\right|=\frac{5}{7} \\ & \Rightarrow \quad\left|\frac{-1}{\sqrt{\left(\frac{2}{5}\right)^2+\frac{1}{b^2}+\frac{1}{c^2}}}\right|=\frac{5}{7} \\ & \Rightarrow \quad(-1)^2\left(\frac{7}{5}\right)^2=\left(\frac{2}{5}\right)^2+\frac{1}{b^2}+\frac{1}{c^2} \\ & \Rightarrow \quad \frac{49}{25}-\frac{4}{25}=\frac{1}{b^2}+\frac{1}{c^2} \\ & \Rightarrow \quad \frac{1}{b^2}+\frac{1}{c^2}=\frac{45}{25}=\frac{9}{5} \quad \text{... (iii)}\\ \end{aligned}$

On squaring Eq. (ii), we get $ \begin{aligned} & \frac{1}{b^2}+\frac{1}{c^2}+\frac{2}{b c}=\frac{9}{25} \\ & \Rightarrow \quad \frac{9}{5}+\frac{2}{b c}=\frac{9}{25} \quad \text{[from Eq. (iii)]}\\ & \Rightarrow \quad \frac{2}{b c}=\frac{-36}{25} \\ & \Rightarrow b c=-\frac{25}{18} \end{aligned}$

Putting $c=-\frac{25}{18} \cdot \frac{1}{b}$ into Eq. (ii), we get

$\frac{1}{b}+\frac{1}{\left(\frac{-25}{18} \cdot \frac{1}{b}\right)}=\frac{3}{5}$

$\begin{array}{lrl} \Rightarrow & \frac{1}{b}-\frac{18 b}{25} =\frac{3}{5} \\ \Rightarrow & 25-18 b^2 =15 b \\ \Rightarrow & 18 b^2+15 b-25=0 \\ \Rightarrow & 18 b^2+30 b-15 b-25=0 \end{array}$

$\begin{array}{lrl} \Rightarrow & 6 b(3 b+5)-5(3 b+5) =0 \\ \Rightarrow & (3 b+5)(6 b-5) =0 \\ \Rightarrow & b =-\frac{5}{3}, \frac{5}{6} \end{array}$

$\because y$-intercept is negative.

$b=-\frac{5}{3}$

Given that, equation of plane parallel to vectors $2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}$ and $\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}$. Also, it passes through $3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}$.

Let the plane be

$a(x-3)+b(y-2)+c(z-6)=0$ ...... (i)

such that,

$\mathrm{t} a(2)+b(\mathrm{l})+c(\mathrm{l}) =0$

$\Rightarrow 2a+b+c =0$ ....... (ii)

and $a(\mathrm{l})+b(-\mathrm{l})+c(\mathrm{l}) =0$

$\Rightarrow \quad a-b+c=0$ ....... (iii)

From Eq. (ii) and (iii)

$\Rightarrow \quad \frac{a}{2}=\frac{b}{-1}=\frac{c}{-3}=\lambda \quad (\text{say})$

$\Rightarrow a=2 \lambda, b=-\lambda, c=-3 \lambda$

From Eq. (i), we get

$\begin{array}{ccrl} & 2 \lambda(x-3)-\lambda(y-2)-3 \lambda(z-6)=0 \\ \Rightarrow & 2(x-3)-(y-2)-3(z-6)=0 \\ \Rightarrow & 2 x-y-3 z=-14 \end{array}$

Given point, $A(-2,4,-5)$ and $B(1,2,3)$

$A B=3 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+8 \hat{\mathbf{k}}$

DR's are $(3,-2,8)$

Then, direction cosines $=(l, m, n)$

where, $l=\frac{3}{\sqrt{3^2+(-2)^2+8^2}}=\frac{3}{\sqrt{77}}$

$\begin{array}{r} m=\frac{(-2)}{\sqrt{77}} \text { and } n=\frac{8}{\sqrt{77}} \\ \therefore \text { DC's are }\left(\frac{3}{\sqrt{77}}, \frac{-2}{\sqrt{77}}, \frac{8}{\sqrt{77}}\right) \end{array}$

Given, points $(2,3,4),(-1,-2,1)$ and $(5,8,7)$

If $\Delta=0$, then points are collinear.

$\Delta \neq 0$, then it represent vertices of triangle

$\begin{aligned} & \Delta=\left|\begin{array}{ccc} 2 & 3 & 4 \\ -1 & -2 & 1 \\ 5 & 8 & 7 \end{array}\right| \\ & =2(-14-8)-3(-7-5)+4(-8+10) \\ & =-44+36+8=0\end{aligned}$

$\Rightarrow$ Given points are collinear.

Given equation of plane is

$4x+3y+2z=2$ ....... (i)

General form is $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$, where $a, b$ and $c$ are intercept on coordinate axis, convert Eq. (i) in general form as

$\frac{4}{2} x+\frac{3}{2} y+\frac{2}{2} z=1$

or $\quad 2 x+\frac{3}{2} y+z=1 \Rightarrow \frac{x}{1 / 2}+\frac{y}{2 / 3}+\frac{z}{1}=1$

$\begin{aligned} & \therefore x \text {-intercept }=\frac{1}{2}=(a) \\ & y \text {-intercept }=\frac{2}{3}=(b) \\ & z \text {-intercept }=1=(c) \end{aligned}$

Then, $a+b+c=\frac{1}{2}+\frac{2}{3}+1=\frac{3+4+6}{6}=\frac{13}{6}$

Since we know that two lines are coplanar if

$\left|\begin{array}{ccc} x_2-x_1 & y_2-y_1 & z_2-z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{array}\right|=0$ .... (i)

Given lines are

$\begin{aligned} L_1: \frac{x-3}{2} & =\frac{y-2}{3}=\frac{z-1}{6} \\ \therefore \quad x_1 & =3, y_1=2, z_1=1 \\ a_1 & =2, b_1=3, c_1=\lambda \\ L_2: \frac{x-z}{3} & =\frac{y-3}{2}=\frac{z-2}{3} \\ a_2 & =3, b_2=2, c_2=3 \end{aligned}$

From Eq. (i)

$\begin{aligned} & \left|\begin{array}{rrr} 1 & -1 & -1 \\ 2 & 3 & \lambda \\ 3 & 2 & 3 \end{array}\right|=0 \Rightarrow\left|\begin{array}{ccc} 1 & 0 & 0 \\ 2 & 5 & \lambda+2 \\ 3 & 5 & 6 \end{array}\right|=0 \\ & \Rightarrow \quad 30-5 \lambda-10=0 \\ & \Rightarrow \quad \lambda=4 \\ & \Rightarrow \sin ^{-1}(\sin 4)+\cos ^{-1}(\cos 4) \\ & (\pi-4)+(2 \pi-4)=(3 \pi-8) \\ & \end{aligned}$

Line passing through $(1,1,-1)$ and parallel to $(1,2,-1)$ is

$\begin{aligned} & L_1: \frac{x-1}{1}=\frac{y-1}{2}=\frac{z+1}{-1}=\lambda \\ & L_2: \frac{x-3}{-1}=\frac{y+2}{5}=\frac{z-2}{-4}=\propto \end{aligned}$

For $A:(\lambda+1,2 \lambda+1,-\lambda-1)$

$\begin{aligned} & \quad=(-\propto+3,5 \propto-2,-4 \propto+2) \\ & \Rightarrow \quad \lambda+\propto=2 \Rightarrow 2 \lambda+1=5 \propto-2 \\ & \Rightarrow \quad 4-2 \propto+1=5 \propto-2 \\ & 1=\propto \quad \text { and } \lambda=1 \\ & A:(2,3,-2) \\ & P: 2 x-y+2 z+7=0 \\ & \Rightarrow 2(\lambda+1)-(2 \lambda+1)+(-2 \lambda-2)+7=0 \\ & \Rightarrow \quad-2 \lambda+6=0 \Rightarrow \lambda=3 \\ & B:(4,7,-4)\\& AB^2=2^2+4^2+2^2=24=2 \sqrt{6} \end{aligned}$

Centroid = (2, 2, 2)

$\because$ Triangle is equilateral

(2, 2, 2) = H = G = S = I

$\therefore$ H + G + S + I = (8, 8, 8)

AB = $\lambda$CD

$\Rightarrow (2,2,3)=\lambda(2,2,k-3)$

$\Rightarrow k-3=3\Rightarrow k=6$

P = Angle bisector

$ = {1 \over 2}\left[ {\left( {{2 \over 3} \pm {5 \over {13}}} \right),\left( {{2 \over 3} \pm {{12} \over {13}}} \right),\left( {{1 \over 3} \pm 0} \right)} \right] = {1 \over 2}\left( {{{41} \over {39}},{{62} \over {39}},{1 \over 3}} \right)$

Parallel to this vector will be (41, 62, 13).

Equation of plane containing line of intersection of the given planes is $(x-2y+z+2)+\lambda(3x-y-z+1)=0$

This plane also passes through (1, 1, 1).

$\therefore (1-2+1+2)+\lambda(3-1-1+1)=0$

$\Rightarrow \lambda=-1$

$\therefore$ Required plane $2x+y-2z-1=0$

$x$-intercept of these plane is (x, 0, 0)

$\Rightarrow 2x-1=0$

$x=\frac{1}{2}$

Given that line L$_1$ is passing through $(2 \hat{\mathbf{i}}-\hat{\mathbf{k}})$ and parallel to $(3 \hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}})$

Let $\mathbf{a}_1=2 \hat{\mathbf{i}}-\hat{\mathbf{k}}$ and $\mathbf{b}_1=3 \hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}}$

equation line $L_1: \mathbf{r}=\mathbf{a}_1+\lambda \mathbf{b}_1$ ..... (i)

Also given line $L_2$ is passing through $(2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-3 \hat{\mathbf{k}})$ and parallel to $(\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}})$

Let $\mathbf{a}_2=2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-3 \hat{\mathbf{k}}$ and $\hat{\mathbf{b}}_2=\hat{\mathbf{i}}-2 \hat{\mathbf{j}}+\hat{\mathbf{k}}$

Equation of line $L_2: \mathbf{r}=\mathbf{a}_2+\mathbf{b}_2$ ..... (ii)

$\therefore$ Shortest distance between $L_1$ and $L_2$

$d=\left|\frac{\left(\mathbf{b}_1 \times \mathbf{b}_2\right) \cdot\left(\mathbf{a}_2-\mathbf{a}_1\right)}{\left|\mathbf{b}_1 \times \mathbf{b}_2\right|}\right|$ ..... (iii)

Consider $\mathbf{b}_1 \times \mathbf{b}_2=\left|\begin{array}{ccc}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 3 & -1 & 2 \\ 1 & -2 & 1\end{array}\right|=3 \hat{\mathbf{i}}-\hat{\mathbf{j}}-5 \hat{\mathbf{k}}$

$\left|\mathbf{b}_1 \times \mathbf{b}_2\right|=\sqrt{3^2+(-1)^2+(-5)^2}=\sqrt{35} $

and $\quad \mathbf{a}_2-\mathbf{a}_1=(2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-3 \hat{\mathbf{k}})-(2 \hat{\mathbf{i}}-\hat{\mathbf{k}})$

$=\hat{\mathbf{j}}-2 \hat{\mathbf{k}}$

$\begin{aligned} \left(\mathbf{b}_1 \times \mathbf{b}_2\right) \cdot\left(\mathbf{a}_2-\mathbf{a}_1\right) & =(3 \hat{\mathbf{i}}-\hat{\mathbf{j}}-5 \hat{\mathbf{k}}) \cdot(\hat{\mathbf{j}}-2 \hat{\mathbf{k}}) \\ & =-1+10=9 \end{aligned}$

Now from Eq. (iii)

$\begin{aligned} d & =\left|\frac{9}{\sqrt{35}}\right| \\ \Rightarrow \quad d & =\frac{9}{\sqrt{35}} \end{aligned}$

In parallelogram, diagonals bisect each other

Let fourth vertex (x, y, z)

Mid-point of $A C=$ mid-point of $B D$

$\begin{aligned} & \left(\frac{2+4}{2}, \frac{4+5}{2}, \frac{-1+1}{2}\right) =\left(\frac{x+3}{2}, \frac{y+6}{2}, \frac{z-1}{2}\right) \\ \Rightarrow \quad & \left(3,\frac{9}{2},0\right) =\left(\frac{x+3}{2}, \frac{y+6}{2}, \frac{z-1}{2}\right) \\ \Rightarrow \quad & \frac{x+3}{2} =3, \frac{y+6}{2}=\frac{9}{2}, \frac{z-1}{2}=0 \\ \Rightarrow \quad & x =3, y=3, z=1 \end{aligned}$

Fourth vertex $(3,3,1)$

$\mathbf{A B}=\mathbf{a}=\mathbf{B}-\mathbf{A}$

$ = (5\widehat i - 6\widehat k) - ( - \widehat i + 2\widehat j - 3\widehat k)$

$ = 6\widehat i - 2\widehat j - 3\widehat k$

$AC = c = C - A$

$ = (4\widehat j - \widehat k) - ( - \widehat i + 2\widehat j - 3\widehat k)$

$ = \widehat i + 2\widehat j + 2\widehat k$

Internal angle bisector $=\hat{\mathbf{c}}+\hat{\mathbf{a}}$

$\begin{aligned} & =\frac{\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}}{\sqrt{1+4+4}}+\frac{6 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}-3 \hat{\mathbf{k}}}{\sqrt{36+4+9}} \\ & =\frac{\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}}{3}+\frac{6 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}-3 \hat{\mathbf{k}}}{7} \\ & =\frac{7 \hat{\mathbf{i}}+18 \hat{\mathbf{i}}}{21}+\frac{14 \hat{\mathbf{j}}-6 \hat{\mathbf{j}}}{21}+\frac{14 \hat{\mathbf{k}}-9 \hat{\mathbf{k}}}{21} \\ & =\frac{25 \hat{\mathbf{i}}}{21}+\frac{8 \hat{\mathbf{j}}}{21}+\frac{5 \hat{\mathbf{k}}}{21} \end{aligned}$

DR's of internal bisector of

$\angle B A C=\frac{25}{21}, \frac{8}{21}, \frac{5}{21}$

Direction Cosines $=\frac{25}{\sqrt{714}}, \frac{8}{\sqrt{714}}, \frac{5}{\sqrt{714}}$

Let $\overline{A B}=a \hat{\mathbf{i}}+b \hat{\mathbf{j}}+c \hat{\mathbf{k}}$

$a, b$ and $c$ are projections of $\overline{A B}$ on $x, y$ and $z$ respectively

Projection of $\overline{A B}$ on $x y$-plane

$=\sqrt{a^2+b^2}=\sqrt{15}$

Projection of $\overline{A B}$ on $y z$-plane

$=\sqrt{b^2+c^2}=\sqrt{46}$

Projection of $\overline{A B}$ on $z x$-plane

$=\sqrt{c^2+a^2}=7$

$\therefore \quad a^2+b^2=15, b^2+c^2=46, c^2+a^2=49$

On solving $a=3, b=\sqrt{6}, c=\sqrt{40}$

Projection of $\overline{A B}$ on $Y$-axis $=b=\sqrt{6}$

* no option matches *

Let Equation of plane passing through $(2,1,3)$ is

$l(x-2)+m(y-1)+n(z-3)=0$ .... (i)

This plane is perpendicular to

$\begin{aligned} x-2 y+2 z+3 & =0 \\ \text {and} \quad 3 x-2 y+4 z-4 & =0 \\ \Rightarrow \quad l-2 m+2 n & =0 \\ \text {and} \quad 3 l-2 m+4 n & =0 \\ \Rightarrow \quad \frac{l}{-8+4} & =\frac{m}{6-4}=\frac{n}{-2+6} \\ \Rightarrow \quad \frac{l}{-4} & =\frac{m}{2}=\frac{n}{4} \Rightarrow \frac{l}{-2}=\frac{m}{1}=\frac{n}{2} \end{aligned}$

From Eq. (i),

$\begin{array}{lr} \therefore & -2(x-2)+(y-1)-2(z-3)=0 \\ \Rightarrow & -2 x+4+y-1+2 z-6=0 \\ \Rightarrow & -2 x+y+2 z-3=0 \\ \Rightarrow & 2 x-y-2 z+3=0 \end{array}$

Let the line joining the points $A(2,4,5)$ and $B(3,5,-4)$ is divided by $Y Z$-plane at $(0, y, z)$ in the ratio $m: n$. Then, by section formula $\frac{2 n+3 m}{m+n}=0$

$\begin{aligned} & \Rightarrow \quad \frac{m}{n}=-\frac{2}{3} \\ & \Rightarrow \quad m: n=2: 3 \end{aligned}$

(externally)

Let the line makes angle $\alpha, \beta, \gamma$ with the coordinate axes, then

$\begin{aligned} & \quad \cos ^2 \alpha+\cos ^2 \beta+\cos ^2 \gamma=1 \\ & \alpha=\beta=\gamma \quad \text { (given) }\\ & \Rightarrow \cos ^2 \alpha+\cos ^2 \alpha+\cos ^2 \alpha=1 \\ & \begin{aligned} \text { or } \end{aligned} \\ & 3 \cos ^2 \alpha=1 \text { or } \cos \alpha= \pm \frac{1}{\sqrt{3}} \\ & \text { Direction cosines are }\left( \pm \frac{1}{\sqrt{3}}, \pm \frac{1}{\sqrt{3}}, \pm \frac{1}{\sqrt{3}}\right) \end{aligned}$

Let $P(x, y, z)$ and $O(0,0,0)$

DR's of $O P=(x, y, z)$

But it is given that DR's of $O P=(1,-2,-2)$

$\because x=1, y=-2, z=-2$

$\therefore$ Coordinate of $P=(1,-2,-2)$