The lines $\frac{x-<!--\; -\; -->2}{2}=\frac{y}{-<!--\; -\; -->2}=\frac{z-<!--\; -\; -->7}{16}$ and $\frac{x+3}{4}=\frac{y+2}{3}=\frac{z+2}{1}$ intersect at the point P. If the distance of P from the line $\frac{x+1}{2}=\frac{y-<!--\; -\; -->1}{3}=\frac{z-<!--\; -\; -->1}{1}$ is $l$, then $14l^2$ is equal to ____

The correct answer is 108.

We are given two lines and must first find their intersection point P, then compute the distance from P to a third line.

Step 1: Write the lines in parametric form.

Line L₁:

$\frac{x-2}{2}=\frac{y}{-2}=\frac{z-7}{16}=t$

So: x = 2 + 2t, y = -2t, z = 7 + 16t, passing through (2, 0, 7) with direction vector d₁ = (2, -2, 16).

Line L₂:

$\frac{x+3}{4}=\frac{y+2}{3}=\frac{z+2}{1}=s$

So: x = -3 + 4s, y = -2 + 3s, z = -2 + s, passing through (-3, -2, -2) with direction vector d₂ = (4, 3, 1).

Step 2: Find the intersection point P.

Setting the parametric equations equal:

From the x-equations: 2 + 2t = -3 + 4s ⟹ 2t - 4s = -5 ...(i)

From the y-equations: -2t = -2 + 3s ⟹ -2t - 3s = -2 ...(ii)

Adding equations (i) and (ii):

(2t - 4s) + (-2t - 3s) = -5 + (-2)

-7s = -7 ⟹ s = 1

Substituting s = 1 into (ii): -2t - 3 = -2 ⟹ -2t = 1 ⟹ t = -1/2

Verification with z-coordinates:
L₁: z = 7 + 16(-1/2) = 7 - 8 = -1
L₂: z = -2 + 1 = -1 ✓

Therefore, the coordinates of P are:

x = 2 + 2(-1/2) = 1, y = -2(-1/2) = 1, z = -1

⟹ P = (1, 1, -1)

Step 3: Find the distance from P to Line L₃.

Line L₃:

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

This line passes through point A = (-1, 1, 1) and has direction vector d = (2, 3, 1).

Form vector AP = P - A = (1 - (-1), 1 - 1, -1 - 1) = (2, 0, -2).

Step 4: Compute the cross product AP × d.

$\overrightarrow{AP}\times \overrightarrow{d}=\left|\begin{array}{ccc}i& j& k\\ 2& 0& -2\\ 2& 3& 1\end{array}\right|$

= i(0·1 - (-2)·3) - j(2·1 - (-2)·2) + k(2·3 - 0·2)
= i(0 + 6) - j(2 + 4) + k(6 - 0)
= (6, -6, 6)

Step 5: Compute magnitudes.

|AP × d| = $\sqrt{6^2+{(-6)}^2+6^2}=\sqrt{108}=6\sqrt{3}$

|d| = $\sqrt{2^2+3^2+1^2}=\sqrt{14}$

Step 6: Calculate the distance l.

$l=\frac{|\overrightarrow{AP}\times \overrightarrow{d}|}{\left|\overrightarrow{d}\right|}=\frac{6\sqrt{3}}{\sqrt{14}}$

Therefore:

$l^2=\frac{36\times 3}{14}=\frac{108}{14}=\frac{54}{7}$

Step 7: Compute 14l².

$14l^2=14\times \frac{54}{7}=2\times 54=108$

Therefore, 14l² = 108.