Q.Name the octant in which each of the following points lies:

We are given two points in space:

$A(\alpha, 4, 7)$ and $B(3, \beta, 8)$.

The YZ‑plane (where $x = 0$) divides segment $AB$ in the ratio $2:3$.

The ZX‑plane (where $y = 0$) divides segment $AB$ in the ratio $4:5$.

We need the point $C$ that divides $AB$ externally in the ratio $\alpha : \beta$.

Concept and intuition

When a coordinate plane divides a segment, it means the point of intersection lies on that plane. For the YZ‑plane, the x‑coordinate of the intersection point is 0. Using the section formula, we can relate the coordinates of A and B to the given ratio. This gives equations to solve for the unknown coordinates $\alpha$ and $\beta$. Once we have $\alpha$ and $\beta$, we apply the external section formula to find C.

Step‑by‑step solution

1. YZ‑plane division (ratio 2:3) The YZ‑plane is $x = 0$. Suppose the point where this plane meets AB divides AB in the ratio $2:3$ (from A to B). Using the section formula for internal division:

2. ZX‑plane division (ratio 4:5) The ZX‑plane is $y = 0$. The point where this plane meets AB divides AB in the ratio $4:5$ (from A to B). Using the section formula:

3. Now we have

The external division ratio is $\alpha : \beta = (-2) : (-5) = 2 : 5$ (since both negative, the ratio is effectively $2:5$ but we must keep sign for external division).

4. External division formula If point C divides AB externally in the ratio $m:n$, then:

Here $m = 2$, $n = 5$ (from $\alpha:\beta = 2:5$).

Compute each coordinate:

So $C = \left( -\frac{16}{3}, 10, \frac{19}{3} \right)$.

5. Match with options This matches option (D).

The key idea: because C and D are on the y‑axis and their vertical distance is fixed, we can parameterise them with a single variable. Then write the equations of lines AC and BD, find their intersection point, and eliminate the parameter to get the relation between $x$ and $y$ that the intersection always satisfies.

1. Set up coordinates for C and D.

   Let C be at $(0, c)$. Since D is above C and CD = 4, D is at $(0, c+4)$.

   (C is below D, so D has the larger y‑coordinate.)

2. Write the equation of line AC.

   A is $(-4, 0)$, C is $(0, c)$.

   Slope $m_{AC} = \frac{c - 0}{0 - (-4)} = \frac{c}{4}$.

   Using point-slope form through A:

3. Write the equation of line BD. B is $(4, 0)$, D is $(0, c+4)$. Slope $m_{BD} = \frac{c+4 - 0}{0 - 4} = -\frac{c+4}{4}$. Through B:

4. Find the intersection point P(x, y) of AC and BD. Equate the two expressions for $y$:

Multiply through by 4:

Expand:

Bring terms together:

5. Find y in terms of c. Substitute $x$ into the equation of AC:

Simplify:

6. Eliminate the parameter c. From $x = \frac{8}{c+2}$, we get $c+2 = \frac{8}{x}$, so $c = \frac{8}{x} - 2 = \frac{8 - 2x}{x}$. Also $c+4 = \frac{8}{x} + 2 = \frac{8 + 2x}{x}$. Substitute into $y = \frac{c(c+4)}{c+2}$:

Notice $(8-2x)(8+2x) = 64 - 4x^2$. So

7. Convert to a standard locus equation. Multiply through by $2x$:

This matches option (B).

The equation $x^2 - 4x + 4y - 8 = 0$ has an $x^2$ term but no $y^2$ term — that's the signature of a parabola that opens sideways (its axis is vertical, opening up or down). To extract all the features asked for, we need to rewrite it in the standard form $(x - h)^2 = 4a(y - k)$.

Let's complete the square in $x$.

1. Complete the square for $x$ Group the $x$ terms: $x^2 - 4x$. Half of $-4$ is $-2$, and $(-2)^2 = 4$. So

Substitute back into the original equation:

2. Isolate the squared term

So the standard form is

Compare with $(x - h)^2 = 4a(y - k)$: here $4a = -4$, so $a = -1$, $h = 2$, $k = 3$. The negative $a$ tells us the parabola opens downward.

3. Vertex

   From the standard form, the vertex is $V = (h, k) = (2, 3)$, which is List-II entry III. So B → III.

4. Focus

   For a parabola $(x - h)^2 = 4a(y - k)$, the focus is at $(h, k + a)$. Here $a = -1$, so focus = $(2, 3 + (-1)) = (2, 2)$. That's List-II entry V (2,2). So A → V.

5. One end of the latus rectum

   The latus rectum is a horizontal line through the focus, length $|4a| = 4$. Its endpoints are at $(h \pm 2a, k + a)$. Since $a = -1$, $2a = -2$, so endpoints are

One end is (4,2), which is List-II entry I. So C → I.

6. Point of intersection of the axis and directrix The axis is the vertical line $x = h = 2$. The directrix is horizontal: $y = k - a = 3 - (-1) = 4$. Their intersection is $(2, 4)$, which is List-II entry IV. So D → IV.

Now we have:

A → V, B → III, C → I, D → IV.

That sequence is V, III, I, IV. Look at the options:

• (A) II III I IV
• (B) IV III I V
• (C) V III IV I
• (D) V III I IV

Option (D) is V III I IV — exactly our mapping.

The problem asks for the point where the plane meets the X-axis — that is, the X-intercept. The intercept form of a plane is $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$, where $a$, $b$, $c$ are the intercepts on the X, Y, Z axes respectively. So if we can write the equation of the plane in this form, the X-intercept $a$ is immediate.

The three given points are not intercepts themselves — they are general points on the plane. So we first find the plane’s equation in standard form $Ax + By + Cz + D = 0$, then convert it to intercept form.

1. Find the plane’s equation.

   Let the plane be $Ax + By + Cz + D = 0$. Since $(2,3,0)$ lies on it:

   $2A + 3B + 0C + D = 0 \quad\Rightarrow\quad 2A + 3B + D = 0$ … (i)

   From $(0,-5,2)$: $0A -5B + 2C + D = 0 \quad\Rightarrow\quad -5B + 2C + D = 0$ … (ii)

   From $(-2,0,3)$: $-2A + 0B + 3C + D = 0 \quad\Rightarrow\quad -2A + 3C + D = 0$ … (iii)

2. Solve for ratios of $A, B, C, D$.

   Subtract (i) from (iii): $(-2A + 3C + D) - (2A + 3B + D) = 0$

   $\Rightarrow -4A - 3B + 3C = 0$ … (iv)

   Subtract (ii) from (iii): $(-2A + 3C + D) - (-5B + 2C + D) = 0$

   $\Rightarrow -2A + 5B + C = 0$ … (v)

   From (v): $C = 2A - 5B$. Substitute into (iv):

   $-4A - 3B + 3(2A - 5B) = 0$

   $\Rightarrow -4A - 3B + 6A - 15B = 0$

   $\Rightarrow 2A - 18B = 0 \quad\Rightarrow\quad A = 9B$.

   Then $C = 2(9B) - 5B = 18B - 5B = 13B$.

   From (i): $2(9B) + 3B + D = 0 \Rightarrow 18B + 3B + D = 0 \Rightarrow 21B + D = 0 \Rightarrow D = -21B$.

   So the plane’s equation is: $9B x + B y + 13B z - 21B = 0$.

   Since $B \neq 0$ (otherwise all coefficients vanish), divide through by $B$:

3. Convert to intercept form. Rearrange: $9x + y + 13z = 21$ Divide both sides by 21:

Simplify the X-intercept: $\frac{21}{9} = \frac{7}{3}$.

So the intercept form is:

Hence the X-intercept is $\frac{7}{3}$, meaning the plane meets the X-axis at $\left(\frac{7}{3}, 0, 0\right)$.

Since $\vec a,\vec b,\vec c$ are non-coplanar, we may equate coefficients.

Line: $\vec r=(1+p)\vec a+2\vec b+(-2p)\vec c.$

Plane: $\vec r=3\vec a-q(\vec c-\vec b)+k(\vec a-\vec b+\vec c)=(3+k)\vec a+(q-k)\vec b+(k-q)\vec c.$

Matching components:

With $p=1$ the intersection point is

Therefore $xyz=(2)(2)(-2)=-8.$

The key idea here is that a plane cutting all three coordinate axes forms a tetrahedron with the coordinate planes. The volume of such a tetrahedron is simply $\frac{1}{6}$ times the product of its intercepts on the axes. Why $\frac{1}{6}$? Because the tetrahedron is one corner of a rectangular box whose sides are the intercept lengths — and a tetrahedron fills exactly one-sixth of that box.

Let's work through it.

1. Find the intercepts. The plane is $3x + 4y - 5z = 60$. To find the $x$-intercept, set $y = 0$ and $z = 0$: $3x = 60$, so $x = 20$. For the $y$-intercept, set $x = 0$ and $z = 0$: $4y = 60$, so $y = 15$. For the $z$-intercept, set $x = 0$ and $y = 0$: $-5z = 60$, so $z = -12$.

2. Apply the volume formula. For a tetrahedron bounded by a plane and the three coordinate planes, the volume is

where $a$, $b$, $c$ are the $x$, $y$, $z$ intercepts respectively. This formula comes from the triple integral $\iiint dV$ over the region, which evaluates to exactly $\frac{|abc|}{6}$.

3. Plug in the numbers. Here $|a| = 20$, $|b| = 15$, $|c| = 12$. So

Compute step by step: $20 \times 15 = 300$, then $300 \times 12 = 3600$. Finally, $\frac{3600}{6} = 600$.