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

### Related Formula textDot Product for Orthogonality: vecA cdot vecB = 0 implies a_1b_1 + a_2b_2 + a_3b_3 = 0$$\text{Dot Product for Orthogonality: } \vec{A} \cdot \vec{B} = 0 \implies a_1b_1 + a_2b_2 + a_3b_3 = 0$$ textDistance between 3D points: d = sqrt(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2$$\text{Distance between 3D points: } d = \sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$$ ### Core Logic Let the first line be L_1: x = y+2 = z = t$L_1: x = y+2 = z = t$. Any point P$P$ on L_1$L_1$ has coordinates (t, t-2, t)$(t, t-2, t)$. Let the second line be L_2: fracx+22 = y = z = s$L_2: \frac{x+2}{2} = y = z = s$. Any point Q$Q$ on L_2$L_2$ has coordinates (2s-2, s, s)$(2s-2, s, s)$. The line segment PQ$PQ$ has direction ratios given by the difference of coordinates: DR_PQ = (2s-2-t, s-(t-2), s-t) = (2s-t-2, s-t+2, s-t)$$DR_{PQ} = (2s-2-t, s-(t-2), s-t) = (2s-t-2, s-t+2, s-t)$$ We are given the fixed direction ratios of PQ$PQ$ as (2, 1, 2)$(2, 1, 2)$. Because direction ratios are proportional, we set up equivalence ratios: frac2s-t-22 = fracs-t+21 = fracs-t2$$\frac{2s-t-2}{2} = \frac{s-t+2}{1} = \frac{s-t}{2}$$ ### Step 1: Solve for Line PQ Using the 2nd and 3rd parts of the proportion: fracs-t+21 = fracs-t2$$\frac{s-t+2}{1} = \frac{s-t}{2}$$ 2s - 2t + 4 = s - t Rightarrow s - t = -4 Rightarrow t = s + 4$$2s - 2t + 4 = s - t \Rightarrow s - t = -4 \Rightarrow t = s + 4$$ Using the 1st and 3rd parts of the proportion: frac2s-t-22 = fracs-t2 Rightarrow 2s-t-2 = s-t Rightarrow s = 2$$\frac{2s-t-2}{2} = \frac{s-t}{2} \Rightarrow 2s-t-2 = s-t \Rightarrow s = 2$$ Substitute s=2$s=2$ to find t$t$: t = 2 + 4 = 6$t = 2 + 4 = 6$ Now, substitute these parameters back to find points P and Q: P = (6, 6-2, 6) = (6, 4, 6)$$P = (6, 6-2, 6) = (6, 4, 6)$$ Q = (2(2)-2, 2, 2) = (2, 2, 2)$$Q = (2(2)-2, 2, 2) = (2, 2, 2)$$ The equation of line PQ$PQ$ passing through Q(2,2,2)$Q(2,2,2)$ with direction ratios (2,1,2)$(2,1,2)$ is: fracx-22 = fracy-21 = fracz-22 = lambda$$\frac{x-2}{2} = \frac{y-2}{1} = \frac{z-2}{2} = \lambda$$

### Step 2: Find Perpendicular Foot F Let F$F$ be the foot of the perpendicular from point A(1, 2, 12)$A(1, 2, 12)$ to the line PQ$PQ$. Any general point on line PQ$PQ$ is F(2lambda+2, lambda+2, 2lambda+2)$F(2\lambda+2, \lambda+2, 2\lambda+2)$. The direction ratios of vector vecAF$\vec{AF}$ are: (2lambda+2-1, lambda+2-2, 2lambda+2-12) = (2lambda+1, lambda, 2lambda-10)$$(2\lambda+2-1, \lambda+2-2, 2\lambda+2-12) = (2\lambda+1, \lambda, 2\lambda-10)$$ Since vecAF$\vec{AF}$ is perpendicular to line PQ$PQ$ (which has direction ratios 2, 1, 2$2, 1, 2$), their dot product must be zero: 2(2lambda+1) + 1(lambda) + 2(2lambda-10) = 0$$2(2\lambda+1) + 1(\lambda) + 2(2\lambda-10) = 0$$ 4lambda + 2 + lambda + 4lambda - 20 = 0$$4\lambda + 2 + \lambda + 4\lambda - 20 = 0$$ 9lambda = 18 Rightarrow lambda = 2$$9\lambda = 18 \Rightarrow \lambda = 2$$ Substitute lambda=2$\lambda=2$ to find the exact coordinates of foot F$F$: F = (2(2)+2, 2+2, 2(2)+2) = (6, 4, 6)$$F = (2(2)+2, 2+2, 2(2)+2) = (6, 4, 6)$$ ### Step 3: Compute Final Distance Squared Calculate the squared distance l^2$l^2$ between A(1, 2, 12)$A(1, 2, 12)$ and F(6, 4, 6)$F(6, 4, 6)$: l^2 = (6-1)^2 + (4-2)^2 + (6-12)^2$$l^2 = (6-1)^2 + (4-2)^2 + (6-12)^2$$ l^2 = 5^2 + 2^2 + (-6)^2$$l^2 = 5^2 + 2^2 + (-6)^2$$ l^2 = 25 + 4 + 36 = 65$$l^2 = 25 + 4 + 36 = 65$$ ### Pattern Recognition Whenever you must link two skew lines with a third intersecting line given constant direction ratios, immediately construct generic parametric points on each skew line. Subtraction yields a vector that is directly proportional to the given constants, instantly solving the system. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Mathematics: Three Dimensional Geometry