Key Concepts and Formulas

1. Direction Vector of a Line: For a line given in vector form $\vec{r} = \vec{a} + t\vec{d}$, the vector $\vec{d}$ is the direction vector of the line. Its components are the direction ratios (DRs).
2. Line Perpendicular to Two Lines: If a line 'l' is perpendicular to two lines $l_1$ and $l_2$, its direction vector $\vec{d_l}$ is parallel to the cross product of the direction vectors of $l_1$ and $l_2$, i.e., $\vec{d_l} \propto \vec{d_1} \times \vec{d_2}$.
3. Intersection of Two Lines: To find the intersection of two lines, express a general point on each line using distinct parameters. Equate their corresponding coordinates and solve the resulting system of equations for the parameters.
4. Distance Formula in 3D: The distance between two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ is given by $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$.
5. First Octant: A point $(x, y, z)$ is in the first octant if all its coordinates are positive, i.e., $x > 0, y > 0, z > 0$.

Step-by-Step Solution

Step 1: Determine Direction Vectors of Lines $l_1$ and $l_2$

We start by identifying the direction vectors of the given lines $l_1$ and $l_2$ from their vector equations. The direction vector is the vector multiplied by the parameter ($t$ or $s$).

For line $l_1$: $\overrightarrow r = (3 + t)\widehat i + ( - 1 + 2t)\widehat j + (4 + 2t)\widehat k$ This can be written as $\overrightarrow r = (3\widehat i - \widehat j + 4\widehat k) + t(\widehat i + 2\widehat j + 2\widehat k)$. The direction vector of $l_1$ is $\vec{d_1} = \widehat i + 2\widehat j + 2\widehat k$.

For line $l_2$: $\overrightarrow r = (3 + 2s)\widehat i + (3 + 2s)\widehat j + (2 + s)\widehat k$ This can be written as $\overrightarrow r = (3\widehat i + 3\widehat j + 2\widehat k) + s(2\widehat i + 2\widehat j + \widehat k)$. The direction vector of $l_2$ is $\vec{d_2} = 2\widehat i + 2\widehat j + \widehat k$.

Step 2: Determine the Direction Vector and Equation of Line 'l'

Line 'l' passes through the origin $(0,0,0)$ and is perpendicular to both $l_1$ and $l_2$. This means its direction vector is parallel to the cross product of $\vec{d_1}$ and $\vec{d_2}$.

Let's calculate the cross product $\vec{d_1} \times \vec{d_2}$: $\vec{d_l} = \vec{d_1} \times \vec{d_2} = \begin{vmatrix} \widehat i & \widehat j & \widehat k \\ 1 & 2 & 2 \\ 2 & 2 & 1 \end{vmatrix}$ $= \widehat i((2)(1) - (2)(2)) - \widehat j((1)(1) - (2)(2)) + \widehat k((1)(2) - (2)(2))$ $= \widehat i(2 - 4) - \widehat j(1 - 4) + \widehat k(2 - 4)$ $= -2\widehat i + 3\widehat j - 2\widehat k$ So, the direction vector of line 'l' is $\vec{d_l} = -2\widehat i + 3\widehat j - 2\widehat k$. Since line 'l' passes through the origin $(0,0,0)$, its vector equation is: $\overrightarrow r = \lambda(-2\widehat i + 3\widehat j - 2\widehat k)$ A general point on line 'l' can be represented as $P_l = (-2\lambda, 3\lambda, -2\lambda)$.

Step 3: Find the Point of Intersection of Line 'l' and Line $l_1$

To find the point of intersection, we equate the coordinates of a general point on line 'l' with a general point on line $l_1$. A general point on $l_1$ is $P_1 = (3+t, -1+2t, 4+2t)$. Equating the coordinates:

1. $-2\lambda = 3+t$
2. $3\lambda = -1+2t$
3. $-2\lambda = 4+2t$

From equations (1) and (3), we have $3+t = 4+2t$. Subtracting $t$ from both sides: $3 = 4+t \implies t = -1$.

Substitute $t = -1$ into equation (1): $-2\lambda = 3 + (-1) \implies -2\lambda = 2 \implies \lambda = -1$.

Now, we verify these values with equation (2): $3\lambda = 3(-1) = -3$. $-1+2t = -1+2(-1) = -1-2 = -3$. Since $-3 = -3$, the values $\lambda=-1$ and $t=-1$ are consistent.

Substitute $\lambda = -1$ into the general point of line 'l' to find the intersection point: $P_{int} = (-2(-1), 3(-1), -2(-1)) = (2, -3, 2)$. This is the point of intersection of 'l' and $l_1$. Let's call this point $P_A = (2, -3, 2)$.

Step 4: Find the General Point on Line $l_2$ and Set up the Distance Equation

A general point on line $l_2$ is $P_B = (3+2s, 3+2s, 2+s)$. We are given that the point $(a,b,c)$ on $l_2$ is at a distance of $\sqrt{17}$ from $P_A(2, -3, 2)$. Using the distance formula: $P_A P_B^2 = ((3+2s)-2)^2 + ((3+2s)-(-3))^2 + ((2+s)-2)^2 = (\sqrt{17})^2$ $(1+2s)^2 + (6+2s)^2 + (s)^2 = 17$

Expand and simplify the equation: $(1 + 4s + 4s^2) + (36 + 24s + 4s^2) + s^2 = 17$ Combine like terms: $9s^2 + 28s + 37 = 17$ $9s^2 + 28s + 20 = 0$

Step 5: Solve for 's' and Identify the Point in the First Octant

We solve the quadratic equation $9s^2 + 28s + 20 = 0$ for $s$ using the quadratic formula $s = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $s = \frac{-28 \pm \sqrt{(28)^2 - 4(9)(20)}}{2(9)}$ $s = \frac{-28 \pm \sqrt{784 - 720}}{18}$ $s = \frac{-28 \pm \sqrt{64}}{18}$ $s = \frac{-28 \pm 8}{18}$ This gives two possible values for $s$:

1. $s_1 = \frac{-28 + 8}{18} = \frac{-20}{18} = -\frac{10}{9}$
2. $s_2 = \frac{-28 - 8}{18} = \frac{-36}{18} = -2$

Now we substitute these values of $s$ back into the general point on $l_2$, $P_B = (3+2s, 3+2s, 2+s)$, and check which point lies in the first octant ($x>0, y>0, z>0$).

For $s = -\frac{10}{9}$: $x = 3 + 2\left(-\frac{10}{9}\right) = 3 - \frac{20}{9} = \frac{27-20}{9} = \frac{7}{9}$ $y = 3 + 2\left(-\frac{10}{9}\right) = 3 - \frac{20}{9} = \frac{7}{9}$ $z = 2 + \left(-\frac{10}{9}\right) = \frac{18-10}{9} = \frac{8}{9}$ The point is $\left(\frac{7}{9}, \frac{7}{9}, \frac{8}{9}\right)$. All coordinates are positive, so this point is in the first octant. This is $(a,b,c)$.

For $s = -2$: $x = 3 + 2(-2) = 3 - 4 = -1$ $y = 3 + 2(-2) = 3 - 4 = -1$ $z = 2 + (-2) = 0$ The point is $(-1, -1, 0)$. This point does not satisfy the first octant condition (x and y coordinates are negative).

Therefore, the coordinates of the point $(a,b,c)$ are $\left(\frac{7}{9}, \frac{7}{9}, \frac{8}{9}\right)$.

Step 6: Calculate $18(a+b+c)$

Now we calculate the required expression $18(a+b+c)$: $a+b+c = \frac{7}{9} + \frac{7}{9} + \frac{8}{9} = \frac{7+7+8}{9} = \frac{22}{9}$ $18(a+b+c) = 18 \times \frac{22}{9}$ $18(a+b+c) = 2 \times 22 = 44$

Common Mistakes & Tips

• Cross Product Calculation: Be very careful with the signs and order of terms when calculating the cross product. A common error is to swap the order of subtraction in the $\widehat j$ component.
• Parameter Consistency: When finding the intersection of two lines, always use different parameters (e.g., $t$ and $s$) for each line to avoid confusion.
• First Octant Condition: Remember that for a point to be in the first octant, all its coordinates must be strictly positive ($x>0, y>0, z>0$). Don't forget to check this condition when multiple solutions arise.

Summary

We began by extracting the direction vectors of lines $l_1$ and $l_2$. Then, we found the direction vector of line 'l' by calculating the cross product of $\vec{d_1}$ and $\vec{d_2}$, as 'l' is perpendicular to both. Using the fact that 'l' passes through the origin, we wrote its equation. We then determined the point of intersection of 'l' and $l_1$. Next, we used the distance formula between this intersection point and a general point on $l_2$ to form a quadratic equation in terms of the parameter 's'. Solving this quadratic gave two possible values for 's'. We checked which of these values resulted in a point in the first octant. Finally, we calculated the sum of the coordinates of this point and multiplied it by 18 to get the final answer.

The final answer is $\boxed{44}$.