Let $$\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}$$, $$\vec{b} = 2\hat{i} + 3\hat{j} - 5\hat{k}$$ and $$\vec{c} = 3\hat{i} - \hat{j} + \lambda\hat{k}$$ be three vectors. Let $$\vec{r}$$ be a unit vector along $$\vec{b} + \vec{c}$$. If $$\vec{r} \cdot \vec{a} = 3$$, then $$3\lambda$$ is equal to :

We are given $$\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}, \quad \vec{b} = 2\hat{i} + 3\hat{j} - 5\hat{k}, \quad \vec{c} = 3\hat{i} - \hat{j} + \lambda\hat{k}$$.

We denote $$\vec{r}$$ as the unit vector along $$\vec{b} + \vec{c}$$ and use the condition $$\vec{r} \cdot \vec{a} = 3$$.

First, we compute $$\vec{b} + \vec{c}$$:

$$\vec{b}+\vec{c} = (2+3)\hat{i} + (3-1)\hat{j} + (-5+\lambda)\hat{k} = 5\hat{i} + 2\hat{j} + (\lambda-5)\hat{k}$$

Next, its magnitude is

$$|\vec{b} + \vec{c}| = \sqrt{25 + 4 + (\lambda - 5)^2} = \sqrt{29 + (\lambda - 5)^2}$$

Therefore the unit vector is

$$\vec{r} = \frac{5\hat{i} + 2\hat{j} + (\lambda - 5)\hat{k}}{\sqrt{29 + (\lambda - 5)^2}}$$.

Using the dot product condition gives

$$\vec{r} \cdot \vec{a} = \frac{5(1) + 2(2) + (\lambda - 5)(3)}{\sqrt{29 + (\lambda - 5)^2}} = \frac{5 + 4 + 3\lambda - 15}{\sqrt{29 + (\lambda - 5)^2}} = \frac{3\lambda - 6}{\sqrt{29 + (\lambda - 5)^2}} = 3$$

This leads to the equation

$$3\lambda - 6 = 3\sqrt{29 + (\lambda - 5)^2}$$

Dividing both sides by 3 yields

$$\lambda - 2 = \sqrt{29 + (\lambda - 5)^2}$$

Since the square root is non-negative, we require $$\lambda \ge 2$$. Squaring both sides gives

$$ (\lambda - 2)^2 = 29 + (\lambda - 5)^2 $$

Expanding both sides results in

$$ \lambda^2 - 4\lambda + 4 = 29 + \lambda^2 - 10\lambda + 25 $$

which simplifies to

$$ -4\lambda + 4 = 54 - 10\lambda $$

and hence

$$ 6\lambda = 50 $$

so that

$$ \lambda = \frac{25}{3} $$

Therefore,

$$ 3\lambda = 25 $$.

The correct value of $$3\lambda$$ is 25.