3d Geometry — JEE Main practice

1. Write $L_1$ and $L_2$ in parametric form using parameters $t$ and $s$ respectively.

$$L_1: \frac{x-2}{-3}=\frac{y-6}{2}=\frac{z-7}{4}=t,\qquad L_2: \frac{x-4}{2}=\frac{y-3}{1}=\frac{z-5}{3}=s$$

2. So the points $C$ (on $L_1$) and $D$ (on $L_2$) can be expressed as

$$C=(-3 t+2,2 t+6,4 t+7),\qquad D=(2 s+4, s+3,3 s+5)$$

3. The direction ratios of $\overrightarrow{CD}$ are found by subtracting coordinates.

$$\overrightarrow{CD}=(2 s+3 t+2,\ s-2 t-3,\ 3 s-4 t-2)$$

4. Since $L_3$ (and hence $CD$) is parallel to $-3\hat{i}+5\hat{j}+16\hat{k}$, the components of $\overrightarrow{CD}$ must be proportional to $(-3,5,16)$. Equating ratios gives two independent linear equations in $s$ and $t$.

$$13 s+9 t+1=0,\qquad 41 s+36 t+26=0$$

5. Solving this system of two equations simultaneously,

$$s=2,\qquad t=-3$$

6. Substituting back gives the actual coordinates of $C$ and $D$.

$$C=(11,0,-5),\qquad D=(8,5,11)$$

7. Using the distance formula to compute $CD^2$,

$$CD^2=(11-8)^2+(0-5)^2+(-5-11)^2=9+25+256=290$$

Hence, the answer is Option A, $290$.

1. Since the line makes equal angles with all three positive coordinate axes, its direction ratios must be equal, i.e., $(1,1,1)$. So the equation of the line through $(-3,5,2)$ is

$$\frac{x+3}{1}=\frac{y-5}{1}=\frac{z-2}{1}=\lambda$$

2. A general point $R$ on this line, using parameter $\lambda$, is

$$R(\lambda-3, \lambda+5, \lambda+2)$$

3. Let $P(-2,r,1)$ be the external point. If $R$ is the foot of perpendicular from $P$ to the line, then $\overrightarrow{PR}$ must be perpendicular to the line's direction vector $(1,1,1)$.

$$\overrightarrow{PR}=(\lambda-1, \lambda+5-r, \lambda+1)$$ $$(\lambda-1)(1)+(\lambda+5-r)(1)+(\lambda+1)(1)=0$$

4. Simplifying,

$$3 \lambda+5-r=0 \Rightarrow \lambda=\frac{r-5}{3}$$

5. Substitute this back into $R$'s coordinates to express $R$ purely in terms of $r$.

$$R \equiv\left(\frac{r-14}{3}, \frac{r+10}{3}, \frac{r+1}{3}\right)$$

6. Since $PR=\sqrt{\frac{14}{3}}$, apply the distance formula between $P(-2,r,1)$ and $R$.

$$\left(\frac{r-14}{3}+2\right)^2+\left(\frac{r+10}{3}-r\right)^2+\left(\frac{r+1}{3}-1\right)^2=\frac{14}{3}$$

7. Simplifying this equation leads to a quadratic in $r$.

$$6 r^2-60 r+126=0 \Rightarrow r^2-10 r+21=0$$

8. Solving,

$$r=3 \text{ or } r=7$$

9. The sum of these possible values is

$$3+7=10$$

Hence, the answer is Option D, $10$.

1. Since $Q(5,b,c)$ is the image of $P(1,2,a)$ in the given line, the segment $PQ$ must be perpendicular to the line's direction vector $(3,2,2)$ (noting $\frac{7-z}{2}$ gives direction ratio $-2$, adjusted sign as per the line's standard form).

$$PQ=(4, b-2, c-a)$$

2. Applying the perpendicularity (dot product equals zero) condition,

$$3(4)+2(b-2)-2(c-a)=0$$

3. Simplifying this equation, along with using the fact that the midpoint of $PQ$ must lie on the given line (which fixes the remaining unknowns), leads to

$$a-c=-12$$

4. Solving the full system (using the midpoint-on-line condition together with the perpendicularity condition) gives

$$a=3,\qquad b=8,\qquad c=15$$

5. Therefore,

$$a^2+b^2+c^2=9+64+225=298$$

Hence, the answer is Option A, $298$.