JEE Mathematics 2025 Question with Solution

Given: The position vectors of the three vertices are $\overrightarrow{p} = 4\hat{i} + \hat{j} - 3\hat{k}$, $\overrightarrow{q} = -5\hat{i} + 2\hat{j} + 3\hat{k}$, and $\overrightarrow{r} = -5\hat{i} + 3\hat{j} + 2\hat{k}$.

Find: The value of $\alpha + 2\beta + 5\gamma$.

From the extracted solution, the correct option is stated to be C and the final value is stated as:

$$\alpha + 2\beta + 5\gamma = 3$$

Therefore, the correct option is C.

Given:

$$\overrightarrow{p} = 4\hat{i} + \hat{j} - 3\hat{k}, \quad \overrightarrow{q} = -5\hat{i} + 2\hat{j} + 3\hat{k}, \quad \overrightarrow{r} = -5\hat{i} + 3\hat{j} + 2\hat{k}$$

Find: $\alpha + 2\beta + 5\gamma$

the solution computes side vectors first:

$$\overrightarrow{pq} = \overrightarrow{q} - \overrightarrow{p} = -9\hat{i} + \hat{j} + 6\hat{k}$$ $$\overrightarrow{pr} = \overrightarrow{r} - \overrightarrow{p} = -9\hat{i} + 2\hat{j} + 5\hat{k}$$

Then it uses the angle formula:

$$\cos \alpha = \frac{\overrightarrow{pq} \cdot \overrightarrow{pr}}{|\overrightarrow{pq}|\,|\overrightarrow{pr}|}$$

with

$$\overrightarrow{pq} \cdot \overrightarrow{pr} = (-9)(-9) + (1)(2) + (6)(5) = 113$$ $$|\overrightarrow{pq}| = \sqrt{118}, \qquad |\overrightarrow{pr}| = \sqrt{110}$$

so

$$\cos \alpha = \frac{113}{\sqrt{118}\sqrt{110}}$$

Similarly, the working gives:

$$\overrightarrow{qp} = 9\hat{i} - \hat{j} - 6\hat{k}, \qquad \overrightarrow{qr} = \hat{j} - \hat{k}$$ $$\cos \beta = \frac{5}{\sqrt{236}}$$

Also,

$$\overrightarrow{rp} = 9\hat{i} - 2\hat{j} - 5\hat{k}, \qquad \overrightarrow{rq} = -\hat{j} + \hat{k}$$ $$\cos \gamma = \frac{-3}{\sqrt{220}}$$

the solution then states that summing the angle measures as required leads to the final answer:

$$\boxed{3}$$

Hence, $\alpha + 2\beta + 5\gamma = 3$, so the correct option is C.

Note: The detailed numerical angle evaluation is not fully established in the provided working, but the source solution explicitly concludes the answer as $3$.