This exercise is a multiple choice questionnaire. For each question, four answers are proposed and only one of them is correct. No justification is required. 1.5 points are awarded for each correct answer. No points are deducted for no answer or an incorrect answer.
Question 1 In space with an orthonormal coordinate system ($\mathrm{O}; \vec{\imath}, \vec{\jmath}, \vec{k}$), we consider the line ($D$) with parametric representation $\left\{\begin{array}{l} x = 2 + t \\ y = 1 - 3t \\ z = 2t \end{array} \quad (t \in \mathbb{R})\right.$, and the plane $(P)$ with Cartesian equation $x + y + z - 3 = 0$.
We can assert that: Answer A: the line ($D$) and the plane ($P$) are strictly parallel. Answer B: the line ($D$) is contained in the plane ($P$). Answer C: the line ($D$) and the plane ($P$) intersect at the point with coordinates $(4; -5; 4)$. Answer D: the line ($D$) and the plane ($P$) are orthogonal.
Question 2 In the computer department of a large store, only one salesperson is present and there are many customers. We assume that the random variable $T$, which associates to each customer the waiting time in minutes for the salesperson to be available, follows an exponential distribution with parameter $\lambda$. The average waiting time is 20 minutes. Given that a customer has already waited 20 minutes, the probability that their total waiting time exceeds half an hour is: Answer A: $\mathrm{e}^{-\frac{1}{2}}$ Answer B: $\mathrm{e}^{-\frac{3}{2}}$ Answer C: $1 - \mathrm{e}^{-\frac{1}{2}}$ Answer D: $1 - \mathrm{e}^{-10\lambda}$
Question 3
A factory manufactures tennis balls in large quantities. To comply with international competition regulations, the diameter of a ball must be between $63.5\text{ mm}$ and $66.7\text{ mm}$. We denote by $D$ the random variable which associates to each ball produced its diameter measured in millimetres. We assume that $D$ follows a normal distribution with mean 65.1 and standard deviation $\sigma$. We call $P$ the probability that a ball chosen at random from the total production is compliant. The factory decides to adjust the machines so that $P$ equals 0.99. The value of $\sigma$, rounded to the nearest hundredth, allowing this objective to be achieved is: Answer A: 0.69 Answer B: 2.58 Answer C: 0.62 Answer D: 0.80
Question 4 The curve below is the graph, in an orthonormal coordinate system, of the function $f$ defined by: $$f(x) = \frac{4x}{x^{2} + 1}$$ The exact value of the positive real number $a$ such that the line with equation $x = a$ divides the shaded region into two regions of equal area is: Answer A: $\sqrt{\sqrt{\frac{3}{2}}}$ Answer B: $\sqrt{\sqrt{5} - 1}$ Answer C: $\ln 5 - 0.5$ Answer D: $\frac{10}{9}$

Let $\alpha$ be the plane passing through point $\mathrm { A } ( 1,2,3 )$ and perpendicular to the line $l : x - 1 = \frac { y - 2 } { - 2 } = \frac { z - 3 } { 3 }$. When the intersection point of plane $\alpha$ and line $m : x - 2 = y = \frac { z - 6 } { 5 }$ is B, what is the length of segment AB? [3 points]
(1) $\sqrt { 19 }$
(2) $\sqrt { 17 }$
(3) $\sqrt { 15 }$
(4) $\sqrt { 13 }$
(5) $\sqrt { 11 }$

In coordinate space, let A be the intersection point of the line $\frac { x } { 2 } = y = z + 3$ and the plane $\alpha : x + 2 y + 2 z = 6$. A sphere with center at point $( 1 , - 1,5 )$ passing through point A intersects plane $\alpha$ to form a figure with area $k \pi$. Find the value of $k$. [3 points]

The borehole is extended in a straight line and leaves the water-bearing rock layer at a depth of 3600 m below the Earth's surface. The exit point is described in the model as a point $R$ on the line $P Q$. Determine the coordinates of $R$ and calculate the thickness of the water-bearing rock layer rounded to the nearest metre. (for verification: $x _ { 1 }$ and $x _ { 2 }$ coordinate of $R : 1.04$ )

Show by calculation that the second borehole reaches the water-bearing rock layer in the model at the point $T ( t | - t | - 4,3 )$, and explain how the length of the second borehole to the water-bearing rock layer is influenced by the location of the associated drilling site.

Water fountains emerge at four points on the surface of the marble sphere. One of these exit points is described in the model by the point $L _ { 0 } ( 1 | 1 | 6 )$. The corresponding fountain is modeled by points $L _ { t } \left( t + 1 | t + 1 | 6,2 - 5 \cdot ( t - 0,2 ) ^ { 2 } \right)$ with suitable values $t \in \mathbb { R } _ { 0 } ^ { + }$.
The point $P$ lies inside the triangle ABS and describes in the model the location where the fountain hits the bronze bowl (see figure). Determine the coordinates of $P$.

Justify by calculation that the skier actually passes through the gate.

155 -- Suppose point $M$ is the intersection of the line passing through points $A(2,2,1)$ and $B(2,-1,5)$ with the plane $x + y + z = 1$. What is the distance from point $M$ to point $B$?

• [(1)] $25$
• [(2)] $27$
• [(3)] $10\sqrt{5}$
• [(4)] $5\sqrt{13}$

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3. In space with orthogonal Cartesian coordinate system $O x y z$, the plane $\pi : 3 x - 2 y + 5 = 0$ is given. －Determine the coordinates of point $H$, the orthogonal projection of $P ( 4,2,1 )$ onto the plane $\pi$; －Determine the intersection of the line $s$ : $\left\{ \begin{array} { l } x - y + 1 = 0 \\ z - 2 = 0 \end{array} \right.$ with the plane $\pi$.

A line with positive direction cosines passes through the point $P(2,-1,2)$ and makes equal angles with the coordinate axes. The line meets the plane $$2x+y+z=9$$ at point $Q$. The length of the line segment $PQ$ equals
(A) 1
(B) $\sqrt{2}$
(C) $\sqrt{3}$
(D) 2

Three lines are given by $$\begin{aligned} & \vec { r } = \lambda \hat { i } , \lambda \in \mathbb { R } \\ & \vec { r } = \mu ( \hat { i } + \hat { j } ) , \quad \mu \in \mathbb { R } \text { and } \\ & \vec { r } = v ( \hat { i } + \hat { j } + \hat { k } ) , \quad v \in \mathbb { R } \end{aligned}$$ Let the lines cut the plane $x + y + z = 1$ at the points $A , B$ and $C$ respectively. If the area of the triangle $A B C$ is $\triangle$ then the value of $( 6 \Delta ) ^ { 2 }$ equals

A straight line drawn from the point $P ( 1,3,2 )$, parallel to the line $\frac { x - 2 } { 1 } = \frac { y - 4 } { 2 } = \frac { z - 6 } { 1 }$, intersects the plane $L _ { 1 } : x - y + 3 z = 6$ at the point $Q$. Another straight line which passes through $Q$ and is perpendicular to the plane $L _ { 1 }$ intersects the plane $L _ { 2 } : 2 x - y + z = - 4$ at the point $R$. Then which of the following statements is (are) TRUE?
(A) The length of the line segment $PQ$ is $\sqrt { 6 }$
(B) The coordinates of $R$ are $( 1,6,3 )$
(C) The centroid of the triangle $PQR$ is $\left( \frac { 4 } { 3 } , \frac { 14 } { 3 } , \frac { 5 } { 3 } \right)$
(D) The perimeter of the triangle $PQR$ is $\sqrt { 2 } + \sqrt { 6 } + \sqrt { 11 }$

The distance of the point $(1, 0, 2)$ from the point of intersection of the line $\frac{x-2}{3} = \frac{y+1}{4} = \frac{z-2}{12}$ and the plane $x - y + z = 16$, is:
(1) $2\sqrt{14}$
(2) $8$
(3) $3\sqrt{21}$
(4) $13$

If the image of the point $P(1, -2, 3)$ in the plane $2x + 3y - 4z + 22 = 0$ measured parallel to the line $\dfrac{x}{1} = \dfrac{y}{4} = \dfrac{z}{5}$ is $Q$, then $PQ$ is equal to:
(1) $3\sqrt{5}$
(2) $2\sqrt{42}$
(3) $\sqrt{42}$
(4) $6\sqrt{5}$

Let $P$ be a plane $l x + m y + n z = 0$ containing the line, $\frac { 1 - x } { 1 } = \frac { y + 4 } { 2 } = \frac { z + 2 } { 3 }$. If plane $P$ divides the line segment $A B$ joining points $A ( - 3 , - 6,1 )$ and $B ( 2,4 , - 3 )$ in ratio $k : 1$ then the value of $k$ is equal to :
(1) 1.5
(2) 3
(3) 2
(4) 4

The square of the distance of the point of intersection of the line $\frac { x - 1 } { 2 } = \frac { y - 2 } { 3 } = \frac { z + 1 } { 6 }$ and the plane $2 x - y + z = 6$ from the point $( - 1 , - 1,2 )$ is

Let the system of linear equations $- x + 2 y - 9 z = 7$ $- x + 3 y + 7 z = 9$ $- 2 x + y + 5 z = 8$ $- 3 x + y + 13 z = \lambda$ has a unique solution $x = \alpha , y = \beta , z = \gamma$. Then the distance of the point $( \alpha , \beta , \gamma )$ from the plane $2 x - 2 y + z = \lambda$ is
(1) 11
(2) 7
(3) 9
(4) 13

Let the image of the point $P(2, -1, 3)$ in the plane $x + 2y - z = 0$ be $Q$. Then the distance of the plane $3x + 2y + z + 29 = 0$ from the point $Q$ is
(1) $\frac{22\sqrt{2}}{7}$
(2) $\frac{24\sqrt{2}}{7}$
(3) $2\sqrt{14}$
(4) $3\sqrt{14}$

Let $P$ be the point of intersection of the line $\frac{x+3}{3} = \frac{y+2}{1} = \frac{1-z}{2}$ and the plane $x + y + z = 2$. If the distance of the point $P$ from the plane $3x - 4y + 12z = 32$ is $q$, then $q$ and $2q$ are the roots of the equation
(1) $x^2 - 18x - 72 = 0$
(2) $x^2 - 18x + 72 = 0$
(3) $x^2 + 18x + 72 = 0$
(4) $x^2 + 18x - 72 = 0$

Let $\mathrm { L } _ { 1 } : \frac { x - 1 } { 3 } = \frac { y - 1 } { - 1 } = \frac { z + 1 } { 0 }$ and $\mathrm { L } _ { 2 } : \frac { x - 2 } { 2 } = \frac { y } { 0 } = \frac { z + 4 } { \alpha } , \alpha \in \mathbf { R }$, be two lines, which intersect at the point $B$. If $P$ is the foot of perpendicular from the point $A ( 1,1 , - 1 )$ on $L _ { 2 }$, then the value of $26 \alpha ( \mathrm {~PB} ) ^ { 2 }$ is $\_\_\_\_$

Let the line passing through the points $(-1, 2, 1)$ and parallel to the line $\frac{x-1}{2} = \frac{y+1}{3} = \frac{z}{4}$ intersect the line $\frac{x+2}{3} = \frac{y-3}{2} = \frac{z-4}{1}$ at the point $P$. Then the distance of $P$ from the point $Q(4, -5, 1)$ is
(1) 5
(2) $5\sqrt{5}$
(3) $5\sqrt{6}$
(4) 10

Let a straight line $L$ pass through the point $P ( 2 , - 1,3 )$ and be perpendicular to the lines $\frac { x - 1 } { 2 } = \frac { y + 1 } { 1 } = \frac { z - 3 } { - 2 }$ and $\frac { x - 3 } { 1 } = \frac { y - 2 } { 3 } = \frac { z + 2 } { 4 }$. If the line $L$ intersects the $y z$-plane at the point $Q$, then the distance between the points $P$ and $Q$ is :
(1) $\sqrt { 10 }$
(2) $2 \sqrt { 3 }$
(3) 2
(4) 3

Q79. Let d be the distance of the point of intersection of the lines $\frac { x + 6 } { 3 } = \frac { y } { 2 } = \frac { z + 1 } { 1 }$ and $\frac { x - 7 } { 4 } = \frac { y - 9 } { 3 } = \frac { z - 4 } { 2 }$ from the point $( 7,8,9 )$. Then $\mathrm { d } ^ { 2 } + 6$ is equal to :
(1) 69
(2) 78
(3) 72
(4) 75

In coordinate space, there is a line $L$ with direction vector $( 1 , - 2, 2 )$, plane $E _ { 1 } : 2 x + 3 y + 6 z = 10$, and plane $E _ { 2 } : 2 x + 3 y + 6 z = - 4$. The length of the line segment of $L$ cut off by $E _ { 1 }$ and $E _ { 2 }$ is . (Express as a fraction in lowest terms)