(A) 360

(B) 348

(C) 320

(D) 340

Correct Answer: 360

Explanation

Step 1: Find intersection point R (verified explicitly)

From $L_1$: $$ R(1+2\lambda,\;2+3\lambda,\;3+4\lambda) $$ From $L_2$: $$ R(4+5\mu,\;1+2\mu,\;\mu) $$

Equating coordinates:

$$ 1+2\lambda = 4+5\mu \quad (1) $$ $$ 2+3\lambda = 1+2\mu \quad (2) $$ $$ 3+4\lambda = \mu \quad (3) $$

From (3): $\mu = 3+4\lambda$

Substitute in (2): $$ 2+3\lambda = 1+2(3+4\lambda) $$ $$ 2+3\lambda = 7+8\lambda $$ $$ -5\lambda = 5 \Rightarrow \lambda = -1 $$

Then from (3): $\mu = 3+4(-1) = -1$

Verification in equation (1):

LHS $= 1+2(-1) = -1$

RHS $= 4+5(-1) = -1$

Hence equation (1) is also satisfied.

Therefore, $$ R=(-1,-1,-1) $$

Step 2: Find point P

Direction vector of $L_1$ is $(2,3,4)$ with magnitude $\sqrt{29}$.

Since $|\overrightarrow{PR}|=\sqrt{29}$ and $P$ lies in first octant:

$$ P=R+(2,3,4)=(1,2,3) $$

Step 3: Find point Q

Let $Q=(4+5t,\;1+2t,\;t)$.

Using $|\overrightarrow{PQ}|^2=\frac{47}{3}$:

$$ (3+5t)^2+(-1+2t)^2+(t-3)^2=\frac{47}{3} $$

$$ 9t^2+6t+1=0 $$

$$ t=-\frac13 $$

$$ Q=\left(\frac73,\frac13,-\frac13\right) $$

Step 4: Compute $27(QR)^2$

$$ (QR)^2=\frac{40}{3} $$

$$ 27(QR)^2=27\times\frac{40}{3}=\boxed{360} $$

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