Distance Between Two Points

The line $$PQ$$ has direction vector

$$\vec d=(1,1,2)$$

Hence,

$$L:\ (x,y,z)=(1,2,-1)+\lambda(1,1,2)$$

Let

$$S=(1+\lambda,2+\lambda,-1+2\lambda)$$

Since $$RS\perp L$$,

$$[(4,-1,5)-S]\cdot(1,1,2)=0$$

Therefore,

$$(3-\lambda)+(-3-\lambda)+2(6-2\lambda)=0$$

$$12-6\lambda=0$$

$$\lambda=2$$

Hence,

$$S=(3,4,3)$$

Now,

$$PS=\sqrt{(2)^2+(2)^2+(4)^2}=2\sqrt6$$

Given,

$$PS:ST=1:2$$

Therefore,

$$ST=4\sqrt6$$

and since $$P,S,T$$ are collinear,

$$S$$ divides $$PT$$ internally in the ratio $$1:2$$.

Using section formula,

$$S=\frac{2P+T}{3}$$

Hence,

$$T=3S-2P$$

$$=(9,12,9)-(2,4,-2)$$

$$=(7,8,11)$$

Now,

$$\overrightarrow{PT}=(6,6,12)=6(1,1,2)$$

and

$$\overrightarrow{RS}=(-1,5,-2)$$

Since

$$\overrightarrow{PT}\cdot\overrightarrow{RS}=0$$

the altitude from $$R$$ passes through $$S$$.

Also,

$$PR^2=(3)^2+(-3)^2+(6)^2=54$$

$$RT^2=(3)^2+(9)^2+(6)^2=126$$

$$PT^2=(6)^2+(6)^2+(12)^2=216$$

Since

$$54+126=180\ne216$$

the triangle is not right-angled.

The area is

$$\frac12|\overrightarrow{PR}\times\overrightarrow{PT}|$$

$$=\frac12|(3,-3,6)\times(6,6,12)|$$

$$=\frac12|(-72,0,36)|$$

$$=\frac12\sqrt{72^2+36^2}$$

$$=\frac12\sqrt{6480}$$

$$=18\sqrt5$$

Hence Statement 4 is FALSE.

To find the orthocentre, the altitude through $$R$$ is

$$R+\mu(-1,5,-2)$$

The altitude through $$P$$ is perpendicular to $$RT$$.

A point common to the two altitudes is

$$\left(\frac{23}{5},-4,\frac{31}{5}\right)$$

Hence Statement 1 is TRUE.

Since line $$L$$ is perpendicular to both $$L_1$$ and $$L_2$$, its direction vector $$\vec{v}$$ is the cross product of the direction vectors of $$L_1$$ and $$L_2$$.

• $$\vec{d_1} = \langle 2, 1, -2 \rangle$$
• $$\vec{d_2} = \langle 1, 3, 4 \rangle$$

$$\vec{v} = \vec{d_1} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -2 \\ 1 & 3 & 4 \end{vmatrix}$$

$$\vec{v} = \hat{i}(4 - (-6)) - \hat{j}(8 - (-2)) + \hat{k}(6 - 1) = \langle 10, -10, 5 \rangle$$.

We can simplify this direction to $$\langle 2, -2, 1 \rangle$$.

The equation of line $$L$$ passing through $$P(2, -1, 3)$$ is:

$$\frac{x-2}{2} = \frac{y+1}{-2} = \frac{z-3}{1} = \lambda$$

General point on $$L$$: $$(2\lambda + 2, -2\lambda - 1, \lambda + 3)$$.

Point $$Q$$ lies on the $$yz$$-plane, where $$x = 0$$:

$$2\lambda + 2 = 0 \implies \lambda = -1$$.

Substitute $$\lambda = -1$$ into the coordinates:

$$Q = (0, -2(-1)-1, -1+3) = \mathbf{(0, 1, 2)}$$.

$$P = (2, -1, 3)$$ and $$Q = (0, 1, 2)$$.

$$PQ = \sqrt{(0-2)^2 + (1-(-1))^2 + (2-3)^2}$$  $$PQ = \sqrt{(-2)^2 + (2)^2 + (-1)^2} = \sqrt{4 + 4 + 1} = \sqrt{9} = \mathbf{3}$$

Any parametric point on the line $$L$$ can be expressed by equating the symmetric equations to a scalar parameter $k$:

$$x = 2k + 1, \quad y = -k + 2, \quad z = k + 3$$

Since point $$A$$ lies on the plane $$P$$, substitute these coordinates into the plane equation $$x + y + z = 6$$:

$$(2k + 1) + (-k + 2) + (k + 3) = 6$$ $$2k + 6 = 6 \implies k = 0$$

Substituting $$k = 0$$ back into our parametric expressions gives the coordinates of point $$A$$:

$$A = (1, 2, 3)$$

The parabola lies in the $$xy$$-plane ($$z = 0$$) and has the equation $$y^2 = 4x$$. We can express any point $$B$$ on this parabola using its standard parametric form $$(t^2, 2t)$$:

$$B = (t^2, 2t, 0)$$

Using the 3D distance formula, write the squared distance $$D^2$$ between $$A(1, 2, 3)$$ and $$B(t^2, 2t, 0)$$:

$$D^2 = (t^2 - 1)^2 + (2t - 2)^2 + (0 - 3)^2$$  $$D^2 = (t^4 - 2t^2 + 1) + (4t^2 - 8t + 4) + 9$$ $$D^2 = t^4 + 2t^2 - 8t + 14$$

To find the minimum value, differentiate $$D^2$$ with respect to $$t$$ and set it to $$0$$:

$$\frac{d(D^2)}{dt} = 4t^3 + 4t - 8 = 0$$ $$t^3 + t - 2 = 0$$

By inspection, $$t = 1$$ is a real root of this cubic equation because $$1^3 + 1 - 2 = 0$$. Factoring out $$(t-1)$$ yields:

$$(t - 1)(t^2 + t + 2) = 0$$

Since the discriminant of the quadratic factor $$t^2 + t + 2$$ is negative ($$1^2 - 4(1)(2) = -7 < 0$$), $$t = 1$$ is the only real critical point.

Substitute $$t = 1$$ back into our distance squared equation:

$$D^2 = (1)^4 + 2(1)^2 - 8(1) + 14 = 1 + 2 - 8 + 14 = 9$$  $$D = \sqrt{9} = 3$$

Correct Option: B