The figure below represents a cube ABCDEFGH. The three points I, J, K are defined by the following conditions:

• I is the midpoint of segment [AD];
• J is such that $\overrightarrow{\mathrm{AJ}} = \frac{3}{4}\overrightarrow{\mathrm{AE}}$;
• K is the midpoint of segment [FG].

1. On the figure provided in the appendix, construct without justification the point of intersection P of the plane (IJK) and the line (EH). Leave the construction lines on the figure.
2. Deduce from this, by justifying, the intersection of the plane (IJK) and the plane (EFG).

Exercise 2 — 5 points Theme: geometry in space Space is equipped with an orthonormal coordinate system $(O; \vec{\imath}, \vec{\jmath}, \vec{k})$. We consider:

• the point $A(1; -1; -1)$;
• the plane $\mathscr{P}_{1}$, with equation: $5x + 2y + 4z = 17$;
• the plane $\mathscr{P}_{2}$ with equation: $10x + 14y + 3z = 19$;
• the line $\mathscr{D}$ with parametric representation: $$\left\{ \begin{aligned} x & = 1 + 2t \\ y & = -t \\ z & = 3 - 2t \end{aligned} \text{ where } t \text{ ranges over } \mathbb{R} . \right.$$

1. Justify that the planes $\mathscr{P}_{1}$ and $\mathscr{P}_{2}$ are not parallel.
2. Prove that $\mathscr{D}$ is the line of intersection of $\mathscr{P}_{1}$ and $\mathscr{P}_{2}$.
3. a. Verify that A does not belong to $\mathscr{P}_{1}$. b. Justify that A does not belong to $\mathscr{D}$.
4. For every real $t$, we denote $M$ the point of $\mathscr{D}$ with coordinates $(1 + 2t; -t; 3 - 2t)$. We then consider the function $f$ which associates to every real $t$ the value $AM^{2}$, that is $f(t) = AM^{2}$. a. Prove that for every real $t$, we have: $f(t) = 9t^{2} - 18t + 17$. b. Prove that the distance AM is minimal when $M$ has coordinates $(3; -1; 1)$.
5. We denote H the point with coordinates $(3; -1; 1)$. Prove that the line (AH) is perpendicular to $\mathscr{D}$.

As shown in the figure, point $N$ is the center of square $ABCD$, $\triangle ECD$ is an equilateral triangle, plane $ECD \perp$ plane $ABCD$, and $M$ is the midpoint of segment $ED$. Then
A. $BM = EN$, and lines $BM$ and $EN$ are intersecting lines
B. $BM \neq EN$, and lines $BM$ and $EN$ are intersecting lines
C. $BM = EN$, and lines $BM$ and $EN$ are skew lines
D. $BM \neq EN$, and lines $BM$ and $EN$ are skew lines

In the cube $A B C D - A _ { 1 } B _ { 1 } C _ { 1 } D _ { 1 }$ , $E , F$ are the midpoints of $A B , B C$ respectively, then
A. Plane $B _ { 1 } E F \perp$ plane $B D D _ { 1 }$
B. Plane $B _ { 1 } E F \perp$ plane $A _ { 1 } B D$
C. Plane $B _ { 1 } E F \parallel$ plane $A _ { 1 } A C$
D. Plane $B _ { 1 } E F \parallel$ plane $A _ { 1 } C _ { 1 } D$

In the cube $ABCD-A_1B_1C_1D_1$, $E, F$ are the midpoints of $AB, BC$ respectively. Then
A. Plane $B_1EF \perp$ plane $BDD_1$
B. Plane $B_1EF \perp$ plane $A_1BD$
C. Plane $B_1EF \parallel$ plane $A_1AC$
D. Plane $B_1EF \parallel$ plane $A_1C_1D$

For all $(t,\theta)\in\mathbb{R}_+\times[0,2\pi]$, define $$F(t,\theta) = \begin{pmatrix} \frac{1}{\sqrt{3}}\operatorname{sh}(t)\cos(\theta) \\ \frac{1}{\sqrt{3}}\operatorname{sh}(t)\sin(\theta) \\ \operatorname{ch}(t) \end{pmatrix}.$$ Show that $F$ takes values in $\mathcal{H}$ and that $F:\mathbb{R}_+\times[0,2\pi]\rightarrow\mathcal{H}$ is surjective.

20. Two planes $\mathrm { P } _ { 1 }$ and $\mathrm { P } _ { 2 }$ pass through origin. Two lines $\mathrm { L } _ { 1 }$ and $\mathrm { L } _ { 2 }$ also passing through origin are such that $\mathrm { L } _ { 1 }$ lies on $\mathrm { P } _ { 1 }$ but not on $\mathrm { P } _ { 2 } , \mathrm {~L} _ { 2 }$ lies on $\mathrm { P } _ { 2 }$ but not on $\mathrm { P } _ { 1 } . \mathrm { A } , \mathrm { B } , \mathrm { C }$ are three points other than origin, then prove that the permutation $\left[ \mathrm { A } ^ { \prime } , \mathrm { B } ^ { \prime } , \mathrm { C } ^ { \prime } \right]$ of $[ \mathrm { A } , \mathrm { B } , \mathrm { C } ]$ exists such that
(i). $\quad \mathrm { A }$ lies on $\mathrm { L } _ { 1 } , \mathrm {~B}$ lies on $\mathrm { P } _ { 1 }$ not on $\mathrm { L } _ { 1 } , \mathrm { C }$ does not lie on $\mathrm { P } _ { 1 }$.
(ii). $\quad \mathrm { A } ^ { \prime }$ lies on $\mathrm { L } _ { 2 } , \mathrm {~B} ^ { \prime }$ lies on $\mathrm { P } _ { 2 }$ not on $\mathrm { L } _ { 2 } , \mathrm { C } ^ { \prime }$ does not lie on $\mathrm { P } _ { 2 }$.
Sol. A corresponds to one of $\mathrm { A } ^ { \prime } , \mathrm { B } ^ { \prime } , \mathrm { C } ^ { \prime }$ and
B corresponds to one of the remaining of $\mathrm { A } ^ { \prime } , \mathrm { B } ^ { \prime } , \mathrm { C } ^ { \prime }$ and
C corresponds to third of $\mathrm { A } ^ { \prime } , \mathrm { B } ^ { \prime } , \mathrm { C } ^ { \prime }$. Hence six such permutations are possible eg One of the permutations may $\mathrm { A } \equiv \mathrm { A } ^ { \prime } ; \mathrm { B } \equiv \mathrm { B } ^ { \prime } , \mathrm { C } \equiv \mathrm { C } ^ { \prime }$ From the given conditions:
A lies on $\mathrm { L } _ { 1 }$.
B lies on the line of intersection of $\mathrm { P } _ { 1 }$ and $\mathrm { P } _ { 2 }$ and ' C ' lies on the line $\mathrm { L } _ { 2 }$ on the plane $\mathrm { P } _ { 2 }$. Now, $\mathrm { A } ^ { \prime }$ lies on $\mathrm { L } _ { 2 } \equiv \mathrm { C }$. $\mathrm { B } ^ { \prime }$ lies on the line of intersection of $\mathrm { P } _ { 1 }$ and $\mathrm { P } _ { 2 } \equiv \mathrm {~B}$ $\mathrm { C } ^ { \prime }$ lie on $\mathrm { L } _ { 1 }$ on plane $\mathrm { P } _ { 1 } \equiv \mathrm {~A}$. Hence there exist a particular set [ $\mathrm { A } ^ { \prime } , \mathrm { B } ^ { \prime } , \mathrm { C } ^ { \prime }$ ] which is the permutation of $[ \mathrm { A } , \mathrm { B } , \mathrm { C } ]$ such that both (i) and
(ii) is satisfied. Here $\left[ \mathrm { A } ^ { \prime } , \mathrm { B } ^ { \prime } , \mathrm { C } ^ { \prime } \right] \equiv [ \mathrm { CBA } ]$.

20. Two planes $\mathrm { P } _ { 1 }$ and $\mathrm { P } _ { 2 }$ pass through origin. Two lines $\mathrm { L } _ { 1 }$ and $\mathrm { L } _ { 2 }$ also passing through origin are such that $\mathrm { L } _ { 1 }$ lies on $\mathrm { P } _ { 1 }$ but not on $\mathrm { P } _ { 2 } , \mathrm {~L} _ { 2 }$ lies on $\mathrm { P } _ { 2 }$ but not on $\mathrm { P } _ { 1 }$. A, B, C are three points other than origin, then prove that the permutation $\left[ \mathrm { A } ^ { \prime } , \mathrm { B } ^ { \prime } , \mathrm { C } ^ { \prime } \right]$ of $[ \mathrm { A } , \mathrm { B } , \mathrm { C } ]$ exists such that
(i). $\quad \mathrm { A }$ lies on $\mathrm { L } _ { 1 } , \mathrm {~B}$ lies on $\mathrm { P } _ { 1 }$ not on $\mathrm { L } _ { 1 } , \mathrm { C }$ does not lie on $\mathrm { P } _ { 1 }$.
(ii). $\quad \mathrm { A } ^ { \prime }$ lies on $\mathrm { L } _ { 2 } , \mathrm {~B} ^ { \prime }$ lies on $\mathrm { P } _ { 2 }$ not on $\mathrm { L } _ { 2 } , \mathrm { C } ^ { \prime }$ does not lie on $\mathrm { P } _ { 2 }$.
Sol. A corresponds to one of $\mathrm { A } ^ { \prime } , \mathrm { B } ^ { \prime } , \mathrm { C } ^ { \prime }$ and
B corresponds to one of the remaining of $\mathrm { A } ^ { \prime } , \mathrm { B } ^ { \prime } , \mathrm { C } ^ { \prime }$ and
C corresponds to third of $\mathrm { A } ^ { \prime } , \mathrm { B } ^ { \prime } , \mathrm { C } ^ { \prime }$. Hence six such permutations are possible eg One of the permutations may $\mathrm { A } \equiv \mathrm { A } ^ { \prime } ; \mathrm { B } \equiv \mathrm { B } ^ { \prime } , \mathrm { C } \equiv \mathrm { C } ^ { \prime }$ From the given conditions:
A lies on $\mathrm { L } _ { 1 }$.
B lies on the line of intersection of $\mathrm { P } _ { 1 }$ and $\mathrm { P } _ { 2 }$ and ' C ' lies on the line $\mathrm { L } _ { 2 }$ on the plane $\mathrm { P } _ { 2 }$. Now, $\mathrm { A } ^ { \prime }$ lies on $\mathrm { L } _ { 2 } \equiv \mathrm { C }$. $\mathrm { B } ^ { \prime }$ lies on the line of intersection of $\mathrm { P } _ { 1 }$ and $\mathrm { P } _ { 2 } \equiv \mathrm {~B}$ $\mathrm { C } ^ { \prime }$ lie on $\mathrm { L } _ { 1 }$ on plane $\mathrm { P } _ { 1 } \equiv \mathrm {~A}$. Hence there exist a particular set [ $\mathrm { A } ^ { \prime } , \mathrm { B } ^ { \prime } , \mathrm { C } ^ { \prime }$ ] which is the permutation of $[ \mathrm { A } , \mathrm { B } , \mathrm { C } ]$ such that both (i) and
(ii) is satisfied. Here $\left[ \mathrm { A } ^ { \prime } , \mathrm { B } ^ { \prime } , \mathrm { C } ^ { \prime } \right] \equiv [ \mathrm { CBA } ]$.

Consider three planes $$\begin{aligned} & P _ { 1 } : x - y + z = 1 \\ & P _ { 2 } : x + y - z = - 1 \\ & P _ { 3 } : x - 3 y + 3 z = 2 . \end{aligned}$$ Let $L _ { 1 } , L _ { 2 } , L _ { 3 }$ be the lines of intersection of the planes $P _ { 2 }$ and $P _ { 3 } , P _ { 3 }$ and $P _ { 1 }$, and $P _ { 1 }$ and $P _ { 2 }$, respectively.
STATEMENT-1: At least two of the lines $L _ { 1 } , L _ { 2 }$ and $L _ { 3 }$ are non-parallel. and STATEMENT-2 : The three planes do not have a common point.
(A) STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is a correct explanation for STATEMENT-1
(B) STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is NOT a correct explanation for STATEMENT-1
(C) STATEMENT-1 is True, STATEMENT-2 is False
(D) STATEMENT-1 is False, STATEMENT-2 is True

The number of distinct real values of $\lambda$, for which the lines $\frac { x - 1 } { 1 } = \frac { y - 2 } { 2 } = \frac { z + 3 } { \lambda ^ { 2 } }$ and $\frac { x - 3 } { 1 } = \frac { y - 2 } { \lambda ^ { 2 } } = \frac { z - 1 } { 2 }$, are coplanar is
(1) 2
(2) 4
(3) 3
(4) 1

A variable plane passes through a fixed point $( 3,2,1 )$ and meets $x , y$ and $z$-axes at $A , B \& C$ respectively. A plane is drawn parallel to the $yz$-plane through $A$, a second plane is drawn parallel to the $zx$-plane through $B$ and a third plane is drawn parallel to the $xy$-plane through $C$. Then the locus of the point of intersection of these three planes, is
(1) $\frac { 3 } { x } + \frac { 2 } { y } + \frac { 1 } { z } = 1$
(2) $\frac { 1 } { x } + \frac { 1 } { y } + \frac { 1 } { z } = \frac { 11 } { 6 }$
(3) $x + y + z = 6$
(4) $\frac { x } { 3 } + \frac { y } { 2 } + \frac { z } { 1 } = 1$

A variable plane passes through a fixed point ( $3,2,1$ ) and meets $x , y$ and $z$ axes at $A , B$ and $C$ respectively. A plane is drawn parallel to $y z$ - plane through $A$, a second plane is drawn parallel $z x$ plane through $B$ and a third plane is drawn parallel to $x y$ - plane through $C$. Then the locus of the point of intersection of these three planes, is
(1) $( x + y + z = 6 )$
(2) $\frac { x } { 3 } + \frac { y } { 2 } + \frac { z } { 1 } = 1$
(3) $\frac { 3 } { x } + \frac { 2 } { y } + \frac { 1 } { z } = 1$
(4) $\frac { 1 } { x } + \frac { 1 } { y } + \frac { 1 } { z } = \frac { 11 } { 6 }$

Let $S$ be the set of all real values of $\lambda$ such that a plane passing through the points $\left( - \lambda ^ { 2 } , 1,1 \right) , \left( 1 , - \lambda ^ { 2 } , 1 \right)$ and $\left( 1,1 , - \lambda ^ { 2 } \right)$ also passes through the point $( - 1 , - 1,1 )$. Then $S$ is equal to :
(1) $\{ \sqrt { 3 } \}$
(2) $\{ 3 , - 3 \}$
(3) $\{ 1 , - 1 \}$
(4) $\{ \sqrt { 3 } , - \sqrt { 3 } \}$

If for some $\alpha$ and $\beta$ in $R$, the intersection of the following three planes $x + 4 y - 2 z = 1$ $x + 7 y - 5 z = \beta$ $x + 5 y + \alpha z = 5$ is a line in $R ^ { 3 }$, then $\alpha + \beta$ is equal to:
(1) 0
(2) 10
(3) 2
(4) - 10

If for some $\alpha \in \mathrm{R}$, the lines $L_1: \frac{x+1}{2} = \frac{y-2}{-1} = \frac{z-1}{1}$ and $L_2: \frac{x+2}{\alpha} = \frac{y+1}{5-\alpha} = \frac{z+1}{1}$ are coplanar, then the line $L_2$ passes through the point:
(1) $(10, 2, 2)$
(2) $(2, -10, -2)$
(3) $(10, -2, -2)$
(4) $(-2, 10, 2)$

The lines $x = ay - 1 = z - 2$ and $x = 3y - 2 = bz - 2 , ( ab \neq 0 )$ are coplanar, if:
(1) $b = 1 , a \in R - \{ 0 \}$
(2) $a = 1 , b \in R - \{ 0 \}$
(3) $a = 2 , b = 2$
(4) $a = 2 , b = 3$

Let the lines $\frac { x - 1 } { \lambda } = \frac { y - 2 } { 1 } = \frac { z - 3 } { 2 }$ and $\frac { x + 26 } { - 2 } = \frac { y + 18 } { 3 } = \frac { z + 28 } { \lambda }$ be coplanar and $P$ be the plane containing these two lines. Then which of the following points does NOT lie on $P$?
(1) $( 0 , - 2 , - 2 )$
(2) $( - 5,0 , - 1 )$
(3) $( 3 , - 1,0 )$
(4) $( 0,4,5 )$

The line, that is coplanar to the line $\frac { x + 3 } { - 3 } = \frac { y - 1 } { 1 } = \frac { z - 5 } { 5 }$, is
(1) $\frac { x + 1 } { - 1 } = \frac { y - 2 } { 2 } = \frac { z - 5 } { 4 }$
(2) $\frac { x + 1 } { - 1 } = \frac { y - 2 } { 2 } = \frac { z - 5 } { 5 }$
(3) $\frac { x - 1 } { - 1 } = \frac { y - 2 } { 2 } = \frac { z - 5 } { 5 }$
(4) $\frac { x + 1 } { 1 } = \frac { y - 2 } { 2 } = \frac { z - 5 } { 5 }$

Let $P$ be the plane passing through the intersection of the planes $\vec{r}\cdot(\hat{i}+3\hat{j}-\hat{k}) = 5$ and $\vec{r}\cdot(2\hat{i}-\hat{j}+\hat{k}) = 3$, and the point $(2, 1, -2)$. Let the position vectors of the points $X$ and $Y$ be $\hat{i} - 2\hat{j} + 4\hat{k}$ and $5\hat{i} - \hat{j} + 2\hat{k}$ respectively. Then the points $X$ and $Y$ with respect to the plane $P$ are
(1) on the same side
(2) on opposite sides
(3) $X$ lies on $P$
(4) $Y$ lies on $P$

Let the lines $L _ { 1 } : \frac { x + 5 } { 3 } = \frac { y + 4 } { 1 } = \frac { z - \alpha } { - 2 }$ and $L _ { 2 } : 3 x + 2 y + z - 2 = 0 = x - 3 y + 2 z - 13$ be coplanar. If the point $P ( a , b , c )$ on $L _ { 1 }$ is nearest to the point $Q ( - 4 , - 3,2 )$, then $| a | + | b | + | c |$ is equal to
(1) 12
(2) 14
(3) 8
(4) 10

In coordinate space, consider three planes $E_{1}: x + y + z = 7$, $E_{2}: x - y + z = 3$, $E_{3}: x - y - z = -5$. Let $L_{3}$ be the line of intersection of $E_{1}$ and $E_{2}$; $L_{1}$ be the line of intersection of $E_{2}$ and $E_{3}$; $L_{2}$ be the line of intersection of $E_{3}$ and $E_{1}$. It is known that the three lines $L_{1}, L_{2}, L_{3}$ have a common intersection point. Find the coordinates of this common intersection point $P$.