3D Geometry

The equation of the plane passing through points $A(-1, -2, -3)$, $B(9, 3, 4)$, and $C(9, -2, 1)$ can be written using the determinant :

Expanding the determinant, we get :

Next, we find the foot of the perpendicular from point $P(3, -2, -9)$ to the plane.

Using the coordinates of $P$ and the equation of the plane, we can find the ratio of the perpendicular distance to the sum of the squares of the coefficients of $x$, $y$, and $z$ :

We can now find the coordinates of the foot of the perpendicular, $Q(\alpha, \beta, \gamma)$ :

Solving for $\alpha$, $\beta$, and $\gamma$ :

So, the coordinates of the foot of the perpendicular are $Q(1, -5, -4)$.

Now, we can find the distance of point $Q$ from the origin :

The equation of the plane is:

We already know that $\lambda = 11$. So the equation of the plane becomes:

Now, let's find the distance $d$ of the point $(\alpha, \beta, \gamma) = (-3, 2, 0)$ from the plane using the formula :

Plugging in the values for the point and the plane equation, we get:

So, the distance of the point $(-3, 2, 0)$ from the plane $2x - 2y + z = 11$ is 7.

Given the two lines :

Let's find the direction vectors of these lines: $\vec{d_1} = \langle 0, 4, 1 \rangle$ and $\vec{d_2} = \langle 3, -4, 0 \rangle$.

Now, let's find the cross product of the direction vectors, which will give a vector that is perpendicular to both lines : $\vec{n} = \vec{d_1} \times \vec{d_2} = \langle 4, 3, -12 \rangle$ Let's find the vector connecting a point on line 1 to a point on line 2 : $\vec{c} = \langle 2\lambda, 3, -12 \rangle$ The shortest distance between the two lines is the projection of $\vec{c}$ onto $\vec{n}$ : $d = \left|\dfrac{\vec{c} \cdot \vec{n}}{|\vec{n}|}\right| = \left|\dfrac{(2\lambda)(4) + (3)(3) - (12)(-12)}{\sqrt{16 + 9 + 144}}\right|$ We are given that the shortest distance is 13 : $13 = \left|\dfrac{8\lambda + 153}{13}\right|$ $|8\lambda + 153| = 169$ We have two cases : 1.

$8\lambda + 153 = 169$ $\lambda = \dfrac{16}{8}$ 2.

$8\lambda + 153 = -169$ $\lambda = \dfrac{-322}{8}$ Now, let's calculate $8\left|\sum\limits_{\lambda \in S} \lambda\right|$ : $8\left|\dfrac{16}{8} + \dfrac{-322}{8}\right| = 8\left|\dfrac{-306}{8}\right| = 306$ Thus, the correct answer is Option A : 306.

Given two lines: $\dfrac{x - x_1}{a_1} = \dfrac{y - y_1}{b_1} = \dfrac{z - z_1}{c_1}$ and $\dfrac{x - x_2}{a_2} = \dfrac{y - y_2}{b_2} = \dfrac{z - z_2}{c_2}$ These lines are coplanar if the determinant of the matrix

= 0 Now let's apply this condition to the given problem.

The given line is : $\dfrac{x + 3}{-3} = \dfrac{y - 1}{1} = \dfrac{z - 5}{5}$ So, the coordinates of any point on this line are $(-3, 1, 5)$ and the direction ratios are $(-3, 1, 5)$.

Now, let's calculate the determinants for each option and check which one equals zero.

For Option A : $\dfrac{x + 1}{-1} = \dfrac{y - 2}{2} = \dfrac{z - 5}{4}$ The coordinates of any point on this line are $(-1, 2, 5)$ and the direction ratios are $(-1, 2, 4)$.

The determinant is :

Applying the formula : = $2(1 \times 4 - 5\times2) - 1((-3\times4) - (5\times-1)) + 0((-3\times2) - (1\times-1))$ = $2(4 - 10) - 1(-12 - (-5)) + 0(-6 - (-1))$ = $2(-6) - 1(-7) + 0(-5)$ = $-12 + 7 + 0$ = $-5$ The determinant for Option A is not equal to zero, so this line is not coplanar with the given line.

For Option B, we have : $\dfrac{x + 1}{-1} = \dfrac{y - 2}{2} = \dfrac{z - 5}{5}$ The coordinates of any point on this line are $(-1, 2, 5)$ and the direction ratios are $(-1, 2, 5)$.

Applying the formula, we get : = $2(1*5 - 5*2) - 1((-3*5) - (5*-1)) + 0((-3*2) - (1*-1))$ = $2(5 - 10) - 1(-15 - (-5)) + 0(-6 - (-1))$ = $2(-5) - 1(-10) + 0(-5)$ = $-10 + 10 + 0$ = $0$ So, the determinant for Option B equals zero, which confirms that the line in Option B is coplanar with the given line.

For Option C : $\dfrac{x - 1}{-1} = \dfrac{y - 2}{2} = \dfrac{z - 5}{5}$ The coordinates of any point on this line are $(1, 2, 5)$ and the direction ratios are $(-1, 2, 5)$.

The determinant is:

Applying the formula : = $4(1*5 - 5*2) - 1((-3*5) - (5*-1)) + 0((-3*2) - (1*-1))$ = $4(5 - 10) - 1(-15 - (-5)) + 0(-6 - (-1))$ = $4(-5) - 1(-10) + 0(-5)$ = $-20 + 10 + 0$ = $-10$ The determinant for Option C is not equal to zero, so this line is not coplanar with the given line.

For Option D : $\dfrac{x + 1}{1} = \dfrac{y - 2}{2} = \dfrac{z - 5}{5}$ The coordinates of any point on this line are $(-1, 2, 5)$ and the direction ratios are $(1, 2, 5)$.

The determinant is :

Applying the formula : = $2(1*5 - 5*2) - 1((-3*5) - (5*1)) + 0((-3*2) - (1*1))$ = $2(5 - 10) - 1(-15 - 5) + 0(-6 - 1)$ = $2(-5) - 1(-20) + 0(-7)$ = $-10 + 20 + 0$ = $10$ So the determinant for Option D is not equal to zero, which means the line in Option D is not coplanar with the given line.

The first step is to find the normal vector to the desired plane.

Since the plane is parallel to the line passing through the points (5, 1, -7) and (1, -1, -1), the direction vector of that line is also parallel to the plane.

The direction vector is the difference between the coordinates of the two points, which is (5-1, 1-(-1), -7-(-1)) = (4, 2, -6).

Next, let's find another vector that is parallel to the plane.

This vector can be obtained by taking the difference between the coordinates of the points (0, -1, 2) and (-1, 2, 1), which the plane passes through.

This gives us a vector of (0-(-1), -1-2, 2-1) = (1, -3, 1).

The normal to the plane is perpendicular to both these vectors.

It can be found by taking the cross product of the two vectors.

The cross product of vectors (4, 2, -6) and (1, -3, 1) is :

The equation of the plane can now be written in the form : -16x - 10y - 14z = d We can find the constant 'd' by substituting one of the points through which the plane passes, say (0, -1, 2) : d = -16.0 -10.(-1) -14.2 = 10 - 28 = -18 So, the equation of the plane is -16x - 10y - 14z = -18.

Now, we substitute the given options into the equation to check which one satisfies it : Option A: (-16.0 -10.5 -14.(-2) = -18) => -50 + 28 = -22 ≠ -18, so A is not correct.

Option B: (-16.2 -10.0 -14.1 = -18) => -32 - 14 = -46 ≠ -18, so B is not correct.

Option C: (-16.1 -10.(-2) -14.1 = -18) => -16 + 20 -14 = -10 ≠ -18, so C is not correct.

Option D: (-16.(-2) -10.5 -14.0 = -18) => 32 - 50 = -18, so D is correct.

So, the plane also passes through the point given in Option D, which is (-2, 5, 0).

Equation of line

Let point $\mathrm{N}(3 \lambda+4,12 \lambda+5,3 \lambda+8)$

And parallel vector to line (say $\vec{a}=3 \hat{i}+12 \hat{j}+3 \hat{k}$ ) Now, $\overrightarrow{\mathrm{PN}} \cdot \overrightarrow{\mathrm{a}}=0$

So point $\mathrm{N}$ is $(2,-3,6)$ Now distance of N(2, -3, 6) from

is $=\left|\dfrac{2(2)-2(-3)+6+5}{\sqrt{4+4+1}}\right|=7$

Given the equation of the plane passing through the intersection of the two given planes:

This is the same as the given equation

. Now, comparing the coefficients of the corresponding variables in both equations:

Solving for $\lambda$:

Now, substituting the value of $\lambda$ back into the equations:

From this equation, we find the value of a :

Now, substituting the value of

and $\lambda$ into the equation:

Therefore, the point

. The given distance is

. The plane is:

. Now, let's find the distance:

Since

, we have

. Therefore,

.

1. Determine the line of shortest distance between the given two lines: Direction vector of line 1:

Direction vector of line 2:

Now, let's find the cross product

So, the direction vector of the line of shortest distance is

. 2. Find the equation of a line passing through point

and having the direction vector

:

3. Find the point

where the line intersects the given plane: Let the coordinates of point

be in terms of $\lambda$:

Since

lies on the plane

, we can substitute the coordinates of

in terms of $\lambda$ into the equation of the plane:

4. Calculate the distance between points

and

: