If the image of the point P (1, 2, a) in the line fracx - 63 = fracy - 72 = frac7 - z2$\frac{x - 6}{3} = \frac{y - 7}{2} = \frac{7 - z}{2}$ is Q(5, b, c), then a^2 + b^2 + c^2$a^2 + b^2 + c^2$ is equal to

### Related Formula textMidpoint M = left(fracx_1 + x_22, fracy_1 + y_22, fracz_1 + z_22right) text lies on the given line.$$\text{Midpoint } M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}, \frac{z_1 + z_2}{2}\right) \text{ lies on the given line.}$$ textDirection ratio of PQ text is perpendicular to line's direction vector vecb, text so overrightarrowPQ cdot vecb = 0.$$\text{Direction ratio of } PQ \text{ is perpendicular to line's direction vector } \vec{b}, \text{ so } \overrightarrow{PQ} \cdot \vec{b} = 0.$$ ### Core Logic Given line L$L$: fracx - 63 = fracy - 72 = fracz - 7-2$\frac{x - 6}{3} = \frac{y - 7}{2} = \frac{z - 7}{-2}$. Notice the standard form requires (z-7)/(-2)$(z-7)/(-2)$. Point P = (1, 2, a)$P = (1, 2, a)$ and its image is Q = (5, b, c)$Q = (5, b, c)$. The midpoint M$M$ of PQ$PQ$ must lie exactly on the line L$L$. M = left(frac1 + 52, frac2 + b2, fraca + c2right) = left(3, fracb + 22, fracc + a2right)$$M = \left(\frac{1 + 5}{2}, \frac{2 + b}{2}, \frac{a + c}{2}\right) = \left(3, \frac{b + 2}{2}, \frac{c + a}{2}\right)$$ ### Step 1: Using the Midpoint on the Line Substitute M$M$ into the line equation: frac3 - 63 = fracfracb + 22 - 72 = fracfracc + a2 - 7-2$$\frac{3 - 6}{3} = \frac{\frac{b + 2}{2} - 7}{2} = \frac{\frac{c + a}{2} - 7}{-2}$$ -1 = fracb - 124 = fracc + a - 14-4$$-1 = \frac{b - 12}{4} = \frac{c + a - 14}{-4}$$ From -1 = fracb - 124$-1 = \frac{b - 12}{4}$, we get -4 = b - 12 implies b = 8$-4 = b - 12 \implies b = 8$. From -1 = fracc + a - 14-4$-1 = \frac{c + a - 14}{-4}$, we get 4 = c + a - 14 implies c + a = 18$4 = c + a - 14 \implies c + a = 18$. ### Step 2: Using the Orthogonality Condition The vector overrightarrowPQ$\overrightarrow{PQ}$ must be perpendicular to the line's direction vector vecv = 3hati + 2hatj - 2hatk$\vec{v} = 3\hat{i} + 2\hat{j} - 2\hat{k}$. overrightarrowPQ = (5 - 1)hati + (b - 2)hatj + (c - a)hatk = 4hati + 6hatj + (c - a)hatk$$\overrightarrow{PQ} = (5 - 1)\hat{i} + (b - 2)\hat{j} + (c - a)\hat{k} = 4\hat{i} + 6\hat{j} + (c - a)\hat{k}$$ (since b = 8$b = 8$) Now set the dot product to zero: 4(3) + 6(2) + (c - a)(-2) = 0$$4(3) + 6(2) + (c - a)(-2) = 0$$ 12 + 12 - 2(c - a) = 0 implies 24 = 2(c - a) implies c - a = 12$$12 + 12 - 2(c - a) = 0 \implies 24 = 2(c - a) \implies c - a = 12$$ ### Step 3: Solving for variables We have a system of linear equations: 1) c + a = 18$c + a = 18$ 2) c - a = 12$c - a = 12$ Adding both: 2c = 30 implies c = 15$2c = 30 \implies c = 15$. Substituting c$c$: 15 + a = 18 implies a = 3$15 + a = 18 \implies a = 3$. Therefore, a = 3, b = 8, c = 15$a = 3, b = 8, c = 15$. Calculate a^2 + b^2 + c^2$a^2 + b^2 + c^2$: a^2 + b^2 + c^2 = 3^2 + 8^2 + 15^2 = 9 + 64 + 225 = 298$$a^2 + b^2 + c^2 = 3^2 + 8^2 + 15^2 = 9 + 64 + 225 = 298$$

### Pattern Recognition Any 'image of a point' problem revolves around two strict constraints: 1) The line bisects the segment joining the point and its image (Midpoint lies on the line), and 2) The segment is orthogonal to the line (Dot product = 0). Directly imposing these generates decoupled simple linear equations. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Three Dimensional Geometry