Key Concepts and Formulas

1. Area of a Triangle using Vectors: The area of a triangle formed by two vectors $\vec{u}$ and $\vec{v}$ originating from the same vertex is given by half the magnitude of their cross product: $\text{Area} = \frac{1}{2} |\vec{u} \times \vec{v}|$
2. Cross Product of Two Vectors: For $\vec{u} = u_x\hat{i} + u_y\hat{j} + u_z\hat{k}$ and $\vec{v} = v_x\hat{i} + v_y\hat{j} + v_z\hat{k}$, their cross product is: $\vec{u} \times \vec{v} = (u_y v_z - u_z v_y)\hat{i} - (u_x v_z - u_z v_x)\hat{j} + (u_x v_y - u_y v_x)\hat{k}$
3. Magnitude of a Vector: The magnitude of a vector $\vec{v} = x\hat{i} + y\hat{j} + z\hat{k}$ is $|\vec{v}| = \sqrt{x^2 + y^2 + z^2}$.
4. Triangle Law of Vector Addition: $\overrightarrow{\mathrm{AB}} + \overrightarrow{\mathrm{BC}} = \overrightarrow{\mathrm{AC}}$.

---

Step-by-Step Solution

Step 1: Calculate the Cross Product $\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}$

We are given $\overrightarrow{\mathrm{AB}}=\hat{i}+2 \hat{j}-7 \hat{k}$ and $\overrightarrow{\mathrm{AC}}=6 \hat{i}+\mathrm{d} \hat{j}-2 \hat{k}$. The area of the triangle ABC can be calculated using the vectors originating from vertex A, i.e., $\overrightarrow{\mathrm{AB}}$ and $\overrightarrow{\mathrm{AC}}$. First, we compute their cross product: $\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -7 \\ 6 & d & -2 \end{vmatrix}$ $= \hat{i}((2)(-2) - (-7)(d)) - \hat{j}((1)(-2) - (-7)(6)) + \hat{k}((1)(d) - (2)(6))$ $= \hat{i}(-4 + 7d) - \hat{j}(-2 + 42) + \hat{k}(d - 12)$ $= (7d - 4)\hat{i} - 40\hat{j} + (d - 12)\hat{k}$

Step 2: Use the Given Area to Determine the Value of 'd'

The area of triangle ABC is given as $15\sqrt{2}$. Using the formula for the area of a triangle formed by vectors: $\text{Area} = \frac{1}{2} |\overrightarrow{\mathrm{AB}} \times \overrightarrow{\mathrm{AC}}|$ $15\sqrt{2} = \frac{1}{2} |(7d - 4)\hat{i} - 40\hat{j} + (d - 12)\hat{k}|$ Multiplying by 2, we get: $30\sqrt{2} = |(7d - 4)\hat{i} - 40\hat{j} + (d - 12)\hat{k}|$ Squaring both sides to eliminate the square root: $(30\sqrt{2})^2 = (7d - 4)^2 + (-40)^2 + (d - 12)^2$ $900 \times 2 = (49d^2 - 56d + 16) + 1600 + (d^2 - 24d + 144)$ $1800 = 50d^2 - 80d + 1760$ Rearranging the terms to form a quadratic equation: $50d^2 - 80d + 1760 - 1800 = 0$ $50d^2 - 80d - 40 = 0$ Dividing the entire equation by 10 for simplification: $5d^2 - 8d - 4 = 0$ Using the quadratic formula $d = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}$ where $A=5$, $B=-8$, $C=-4$: $d = \frac{-(-8) \pm \sqrt{(-8)^2 - 4(5)(-4)}}{2(5)}$ $d = \frac{8 \pm \sqrt{64 + 80}}{10}$ $d = \frac{8 \pm \sqrt{144}}{10}$ $d = \frac{8 \pm 12}{10}$ This gives two possible values for $d$: $d_1 = \frac{8 + 12}{10} = \frac{20}{10} = 2$ $d_2 = \frac{8 - 12}{10} = \frac{-4}{10} = -\frac{2}{5}$ The problem states that $d > 0$, so we choose $d=2$.

Step 3: Determine the Vector $\overrightarrow{\mathrm{BC}}$

With $d=2$, the vector $\overrightarrow{\mathrm{AC}}$ is $6\hat{i} + 2\hat{j} - 2\hat{k}$. Using the triangle law of vector addition, $\overrightarrow{\mathrm{AB}} + \overrightarrow{\mathrm{BC}} = \overrightarrow{\mathrm{AC}}$. We can find $\overrightarrow{\mathrm{BC}}$ by rearranging this equation: $\overrightarrow{\mathrm{BC}} = \overrightarrow{\mathrm{AC}} - \overrightarrow{\mathrm{AB}}$ Substitute the given and calculated vector components: $\overrightarrow{\mathrm{BC}} = (6\hat{i} + 2\hat{j} - 2\hat{k}) - (\hat{i} + 2\hat{j} - 7\hat{k})$ $\overrightarrow{\mathrm{BC}} = (6-1)\hat{i} + (2-2)\hat{j} + (-2 - (-7))\hat{k}$ $\overrightarrow{\mathrm{BC}} = 5\hat{i} + 0\hat{j} + 5\hat{k}$ So, $\overrightarrow{\mathrm{BC}} = 5\hat{i} + 5\hat{k}$.

Step 4: Calculate the Square of the Lengths of All Sides

We now have all three side vectors: $\overrightarrow{\mathrm{AB}} = \hat{i} + 2\hat{j} - 7\hat{k}$ $\overrightarrow{\mathrm{AC}} = 6\hat{i} + 2\hat{j} - 2\hat{k}$ $\overrightarrow{\mathrm{BC}} = 5\hat{i} + 0\hat{j} + 5\hat{k}$

We need to find the square of the length of each side:

1. Square of the length of AB: $|\overrightarrow{\mathrm{AB}}|^2 = (1)^2 + (2)^2 + (-7)^2 = 1 + 4 + 49 = 54$
2. Square of the length of AC: $|\overrightarrow{\mathrm{AC}}|^2 = (6)^2 + (2)^2 + (-2)^2 = 36 + 4 + 4 = 44$
3. Square of the length of BC: $|\overrightarrow{\mathrm{BC}}|^2 = (5)^2 + (0)^2 + (5)^2 = 25 + 0 + 25 = 50$

Step 5: Identify the Largest Side and its Square Length

Comparing the squares of the lengths of the sides: $|\overrightarrow{\mathrm{AB}}|^2 = 54$ $|\overrightarrow{\mathrm{AC}}|^2 = 44$ $|\overrightarrow{\mathrm{BC}}|^2 = 50$

The largest value among these is $54$. Therefore, the largest side is AB, and the square of its length is $54$.

---

Common Mistakes & Tips

• Algebraic Errors: Be meticulous with algebraic manipulations, especially when expanding squared terms and solving quadratic equations.
• Sign Errors in Cross Product: The middle term in the cross product determinant has a negative sign. Ensure this is handled correctly.
• Using the Constraint: Always remember to use the given constraints, such as $d>0$, to select the correct solution for variables.
• Square vs. Length: The question asks for the square of the length, so ensure you are comparing the squared magnitudes, not the magnitudes themselves.

---

Summary

The problem was solved by first utilizing the given area of the triangle to establish a relationship involving the unknown component $d$ through the cross product of two sides. Solving the resulting quadratic equation, and applying the constraint $d>0$, we found $d=2$. Subsequently, the triangle law of vector addition was used to determine the vector representing the third side of the triangle. Finally, the squares of the lengths of all three sides were calculated, and the largest among them was identified.

The final answer is $\boxed{54}$.