Let the line L$L$ pass through the point (-3, 5, 2)$(-3, 5, 2)$ and make equal angles with the positive coordinate axes. If the distance of L$L$ from the point (-2, r, 1)$(-2, r, 1)$ is sqrtfrac143$\sqrt{\frac{14}{3}}$, then the sum of all possible values of r$r$ is:

### Related Formula textEquation of a line: fracx - x_1a = fracy - y_1b = fracz - z_1c = lambda$$\text{Equation of a line: } \frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c} = \lambda$$ textDistance formula: d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2$$\text{Distance formula: } d^2 = (x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2$$ ### Core Logic

Line L$L$ makes equal angles with the positive axes, so its direction ratios are (1, 1, 1)$(1, 1, 1)$. Equation of line L$L$ is fracx + 31 = fracy - 51 = fracz - 21 = lambda$\frac{x + 3}{1} = \frac{y - 5}{1} = \frac{z - 2}{1} = \lambda$. A general point R$R$ on the line is (lambda - 3, lambda + 5, lambda + 2)$(\lambda - 3, \lambda + 5, \lambda + 2)$. Use the perpendicularity condition overrightarrowPR cdot vecd = 0$\overrightarrow{PR} \cdot \vec{d} = 0$ to find lambda$\lambda$, then substitute back into the distance formula. ### Step 1: Apply Perpendicularity Condition Let point P = (-2, r, 1)$P = (-2, r, 1)$. The vector overrightarrowPR = langle lambda - 1, lambda + 5 - r, lambda + 1 rangle$\overrightarrow{PR} = \langle \lambda - 1, \lambda + 5 - r, \lambda + 1 \rangle$. Since overrightarrowPR cdot langle 1, 1, 1 rangle = 0$\overrightarrow{PR} \cdot \langle 1, 1, 1 \rangle = 0$: (lambda - 1)(1) + (lambda + 5 - r)(1) + (lambda + 1)(1) = 0$$(\lambda - 1)(1) + (\lambda + 5 - r)(1) + (\lambda + 1)(1) = 0$$ 3lambda - r + 5 = 0 implies lambda = fracr - 53$$3\lambda - r + 5 = 0 \implies \lambda = \frac{r - 5}{3}$$ ### Step 2: Calculate Distance Substitute lambda$\lambda$ back into point R$R$: R equiv left( fracr - 143, fracr + 103, fracr + 13 right)$$R \equiv \left( \frac{r - 14}{3}, \frac{r + 10}{3}, \frac{r + 1}{3} \right)$$ We are given PR = sqrtfrac143 implies PR^2 = frac143$PR = \sqrt{\frac{14}{3}} \implies PR^2 = \frac{14}{3}$. PR^2 = left( fracr - 143 + 2 right)^2 + left( fracr + 103 - r right)^2 + left( fracr + 13 - 1 right)^2 = frac143$PR^2 = \left( \frac{r - 14}{3} + 2 \right)^2 + \left( \frac{r + 10}{3} - r \right)^2 + \left( \frac{r + 1}{3} - 1 \right)^2 = \frac{14}{3}$ frac(r - 8)^29 + frac(10 - 2r)^29 + frac(r - 2)^29 = frac143$$\frac{(r - 8)^2}{9} + \frac{(10 - 2r)^2}{9} + \frac{(r - 2)^2}{9} = \frac{14}{3}$$ (r^2 - 16r + 64) + (100 + 4r^2 - 40r) + (r^2 - 4r + 4) = 42$$(r^2 - 16r + 64) + (100 + 4r^2 - 40r) + (r^2 - 4r + 4) = 42$$ 6r^2 - 60r + 168 = 42 implies 6r^2 - 60r + 126 = 0$$6r^2 - 60r + 168 = 42 \implies 6r^2 - 60r + 126 = 0$$ Dividing by 6: r^2 - 10r + 21 = 0$$r^2 - 10r + 21 = 0$$ ### Step 3: Solve for r (r - 7)(r - 3) = 0 implies r = 3, 7$$(r - 7)(r - 3) = 0 \implies r = 3, 7$$ The sum of all possible values of r$r$ is 3 + 7 = 10$3 + 7 = 10$. ### Pattern Recognition When a line makes equal angles with coordinate axes, its direction cosines are (1/sqrt3, 1/sqrt3, 1/sqrt3)$(1/\sqrt{3}, 1/\sqrt{3}, 1/\sqrt{3})$, making its simpler direction ratios (1, 1, 1)$(1, 1, 1)$. Use projection vector methods or direct dot product to find perpendicular foot. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 12 Maths: Three Dimensional Geometry