A line with direction ratio 2, 1, 2 meets the lines x = y + 2 = z and x + 2 = 2y = 2z respectively at the points P and Q. If the length of the perpendicular from the point (1, 2, 12) to the line PQ is $l$, then $l^2$  is _____ .

The correct answer is l² = 65.

We are given a line with direction ratios (2, 1, 2) that meets two lines at points P and Q respectively. Let us first write the two lines in parametric form.

Line 1: x = y + 2 = z

Rewriting: $\frac{x}{1}=\frac{y+2}{1}=\frac{z}{1}=t$

So any point on Line 1 is P = (t, t - 2, t), with direction ratios (1, 1, 1).

Line 2: x + 2 = 2y = 2z

Rewriting: $\frac{x+2}{2}=\frac{y}{1}=\frac{z}{1}=s$

So any point on Line 2 is Q = (2s - 2, s, s), with direction ratios (2, 1, 1).

Step 1: Finding Point P

The line through P = (t, t-2, t) with direction ratios (2, 1, 2) has parametric equations:

$(x,y,z)=(t+2\lambda ,t-2+\lambda ,t+2\lambda )$

For this to meet Line 2 at Q = (2s - 2, s, s), we equate:

(i)   t + 2λ = 2s - 2
(ii)   t - 2 + λ = s
(iii)   t + 2λ = s

From equations (i) and (iii):   2s - 2 = s   ⟹   s = 2

From equation (iii): t + 2λ = 2   ⟹   t = 2 - 2λ

Substituting into (ii): (2 - 2λ) - 2 + λ = 2   ⟹   -λ = 2   ⟹   λ = -2

Therefore: t = 2 - 2(-2) = 6

So: P = (6, 4, 6) and Q = (2(2)-2, 2, 2) = (2, 2, 2)

Verification: Direction of PQ = Q - P = (2-6, 2-4, 2-6) = (-4, -2, -4) = -2(2, 1, 2) ✓ — this confirms the direction ratio (2, 1, 2) is satisfied.

Step 2: Equation of Line PQ

Line PQ passes through P = (6, 4, 6) with direction ratios (2, 1, 2):

$\frac{x-6}{2}=\frac{y-4}{1}=\frac{z-6}{2}=\mu$

A general point on PQ is: (6 + 2μ, 4 + μ, 6 + 2μ)

Step 3: Finding the Foot of the Perpendicular from (1, 2, 12)

The vector from a general point on PQ to (1, 2, 12) is:

$(1-6-2\mu ,2-4-\mu ,12-6-2\mu )=(-5-2\mu ,-2-\mu ,6-2\mu )$

For perpendicularity, this vector must be orthogonal to direction (2, 1, 2):

$2(-5-2\mu )+1(-2-\mu )+2(6-2\mu )=0$

$-10-4\mu -2-\mu +12-4\mu =0$

$0-9\mu =0⟹\mu =0$

So the foot of the perpendicular is the point P itself: (6, 4, 6).

Step 4: Computing the Perpendicular Distance l

$l=\sqrt{(1-6{)}^2+(2-4{)}^2+(12-6{)}^2}$

$l=\sqrt{{(-5)}^2+{(-2)}^2+{\left(6\right)}^2}=\sqrt{25+4+36}=\sqrt{65}$

Therefore:

$l^2=65$