Question

Match the entries of List-I with the correct entries of List-II

	List-I
(P)	If the point on y-axls which is equidistant from (2, 4, - 3) and (-3, 5, 1) Is (α, β, γ), then the value of α + 2β + 3γ Is	(1)
(Q)	If the point In yz plane which is equidistant from (3, 0, 2), (2, 3, 0) and (1, 0, 0) Is (α, β, γ), then the value of α + 2β + 3γ Is	(2)
(R)	If the area of triangle with vertices (5, 4, 7), (-1, 1, 9) and (2, 6, 1) Is Δ square unlts, then the value of 2Δ - 36 Is	(3)
(S)	Let the mid point of line segment $P_1P_2$ Is $(2\sqrt{5}, 4, 3)$ and perpendiculars from $P_1$ and $P_2$ to the xy-plane are $Q_1$ and $Q_2$ respectively. Then the distance of mid-point of $Q_1Q_2$ from origin Is	(4)
		(5) 1

Then the correct option is

• A. (P)→(1), (Q)→(2), (R)→(3), (S)→(4)

Answer 1

Answer: (A)

(P)→(1), (Q)→(2), (R)→(3), (S)→(4)

Answer 2

Solution:

We solve each part.

(P) A point on the y‑axis has coordinates $(0,y,0)$. It is equidistant from $A(2,4,-3)$ and $B(-3,5,1)$. Setting distances equal:

$$(0-2)^2+(y-4)^2+(0+3)^2 = (0+3)^2+(y-5)^2+(0-1)^2.$$

That is:

$$4+(y-4)^2+9 = 9+(y-5)^2+1.$$

So,

$$(y-4)^2+13 = (y-5)^2+10.$$

Expanding,

$$y^2-8y+29 = y^2-10y+35.$$

Thus,

$$2y = 6\quad\Longrightarrow\quad y=3.$$

So the point is $(0,3,0)$ and

$$\alpha+2\beta+3\gamma = 0+2\cdot3+0 = 6.$$

(Q) A point in the $yz$-plane has coordinates $(0,y,z)$. It is equidistant from $A(3,0,2)$, $B(2,3,0)$ and $C(1,0,0)$.

Distance squares are:

$$d_A^2 = (0-3)^2 + (y-0)^2 + (z-2)^2 = 9+y^2+(z-2)^2,$$ $$d_B^2 = (0-2)^2 + (y-3)^2 + (z-0)^2 = 4+(y-3)^2+z^2,$$ $$d_C^2 = (0-1)^2 + (y-0)^2 + (z-0)^2 = 1+y^2+z^2.$$

Equate $d_A^2 = d_C^2$:

$$9+y^2+(z-2)^2 = 1+y^2+z^2.$$

Cancel $y^2$ and simplify:

$$9+ z^2-4z+4 = 1+ z^2\quad\Longrightarrow\quad 13-4z=1\quad\Longrightarrow\quad z=3.$$

Now equate $d_B^2 = d_C^2$:

$$4+(y-3)^2+9 = 1+y^2+9.$$

That is,

$$(y-3)^2+13 = y^2+10.$$

Expanding, $y^2-6y+9+13 = y^2+10$ so:

$$y^2-6y+22 = y^2+10\quad\Longrightarrow\quad -6y = -12\quad\Longrightarrow\quad y=2.$$

So the point is $(0,2,3)$ and

$$\alpha+2\beta+3\gamma = 0+2\cdot2+3\cdot3 = 4+9 = 13.$$

(R) Vertices: $A(5,4,7)$, $B(-1,1,9)$, $C(2,6,1)$. Let

$$\vec{AB} = B-A = (-6,-3,2),\quad \vec{AC} = C-A = (-3,2,-6).$$

The area of the triangle is:

$$\Delta = \frac{1}{2}\,|\vec{AB}\times\vec{AC}|.$$

Compute the cross product:

$$\vec{AB}\times\vec{AC} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k}\\ -6 & -3 & 2\\ -3 & 2 & -6 \end{vmatrix} = \big( (-3)(-6)- (2)(2) \big)\mathbf{i} - \big( (-6)(-6)- (2)(-3) \big)\mathbf{j} + \big( (-6)(2)- (-3)(-3) \big)\mathbf{k}.$$

Calculations:

$$\mathbf{i}: (18-4)=14,\quad \mathbf{j}: (36+6)=42\quad \text{(with minus sign gives }-42\text{)},\quad \mathbf{k}: (-12-9)=-21.$$

So,

$$|\vec{AB}\times\vec{AC}| = \sqrt{14^2+(-42)^2+(-21)^2} = \sqrt{196+1764+441} = \sqrt{2401}=49.$$

Thus,

$$\Delta = \frac{1}{2}\times49 = \frac{49}{2}.$$

Then,

$$2\Delta-36 = 49-36 = 13.$$

(S) The midpoint of $P_1P_2$ is given as $M(2\sqrt5,4,3)$. The perpendicular (projection) of any point onto the $xy$-plane simply drops the $z$-coordinate. Thus, the projections of $P_1$ and $P_2$ have a midpoint which is the projection of $M$, i.e., $(2\sqrt5,4,0)$. Its distance from the origin is:

$$\sqrt{(2\sqrt5)^2+4^2} = \sqrt{20+16} = \sqrt{36}=6.$$

Matching with List‑II: List‑I answers:

• (P) gives 6
• (Q) gives 13
• (R) gives 13
• (S) gives 6

If we assign List‑II options as given in the table rows:

$$(P)\to (1),\quad (Q)\to (2),\quad (R)\to (3),\quad (S)\to (4),$$

and the extra option (5) is 1 (which does not match any), the correct matching is:

$$\boxed{(P)\to (1),\quad (Q)\to (2),\quad (R)\to (3),\quad (S)\to (4)}$$

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Explanation (Minimal):

1. For (P), using a point on the y‑axis $(0,y,0)$ and equating distances from $(2,4,-3)$ and $(-3,5,1)$ yields $y=3$. Then $\alpha+2\beta+3\gamma=6$.
2. For (Q), with point on the $yz$-plane $(0,y,z)$ and equating distances from three given points yields $y=2$ and $z=3$, so the sum is 13.
3. For (R), computing the area of the triangle using the cross product gives $\Delta=\frac{49}{2}$ and then $2\Delta-36=13$.
4. For (S), the projection of the midpoint $(2\sqrt5,4,3)$ onto the $xy$-plane is $(2\sqrt5,4,0)$ with distance 6 from the origin.