If the distance of the point $$P(43, \alpha, \beta), \beta<0,$$ from the line $$\overrightarrow{r} = 4\widehat{i}-\widehat{k}+\mu(2\widehat{i}+3\widehat{k}), \mu \in \mathbb{R}$$ along a line with direction ratios 3, -1, 0 is $$13\sqrt{10},$$ then $$ \alpha ^{2}+ \beta^{2}$$ is equal to______

The line is given by $$\overrightarrow{r} = 4\widehat{i} - \widehat{k} + \mu(2\widehat{i} + 3\widehat{k})$$, so parametric equations for any point on the line are:

$$x = 4 + 2\mu, \quad y = 0, \quad z = -1 + 3\mu$$

Let $$Q(4 + 2\mu, 0, -1 + 3\mu)$$ be a point on the line such that the line segment $$PQ$$ has direction ratios 3, -1, 0. The vector $$\overrightarrow{PQ}$$ is:

$$\overrightarrow{PQ} = (4 + 2\mu - 43, 0 - \alpha, -1 + 3\mu - \beta) = (2\mu - 39, -\alpha, 3\mu - 1 - \beta)$$

Since $$\overrightarrow{PQ}$$ is parallel to the vector $$(3, -1, 0)$$, their components are proportional. The z-component of the direction vector is 0, so the z-component of $$\overrightarrow{PQ}$$ must be 0:

$$3\mu - 1 - \beta = 0 \quad \Rightarrow \quad \beta = 3\mu - 1 \quad \text{(1)}$$

The proportionality between x and y components gives:

$$\frac{2\mu - 39}{3} = \frac{-\alpha}{-1} = \alpha \quad \Rightarrow \quad \alpha = \frac{2\mu - 39}{3} \quad \text{(2)}$$

The distance $$|\overrightarrow{PQ}|$$ is given as $$13\sqrt{10}$$:

$$|\overrightarrow{PQ}| = \sqrt{(2\mu - 39)^2 + (-\alpha)^2 + 0^2} = \sqrt{(2\mu - 39)^2 + \alpha^2} = 13\sqrt{10}$$

Squaring both sides:

$$(2\mu - 39)^2 + \alpha^2 = (13\sqrt{10})^2 = 169 \times 10 = 1690 \quad \text{(3)}$$

Substitute $$\alpha$$ from equation (2) into equation (3):

$$(2\mu - 39)^2 + \left(\frac{2\mu - 39}{3}\right)^2 = 1690$$

Let $$t = 2\mu - 39$$:

$$t^2 + \left(\frac{t}{3}\right)^2 = 1690 \quad \Rightarrow \quad t^2 + \frac{t^2}{9} = 1690 \quad \Rightarrow \quad \frac{10t^2}{9} = 1690$$

Multiply both sides by 9:

$$10t^2 = 15210 \quad \Rightarrow \quad t^2 = 1521 \quad \Rightarrow \quad t = \pm 39$$

Case 1: $$t = 39$$

$$2\mu - 39 = 39 \quad \Rightarrow \quad 2\mu = 78 \quad \Rightarrow \quad \mu = 39$$

From equation (1): $$\beta = 3(39) - 1 = 117 - 1 = 116$$

But $$\beta < 0$$, and 116 is positive, so this case is invalid.

Case 2: $$t = -39$$

$$2\mu - 39 = -39 \quad \Rightarrow \quad 2\mu = 0 \quad \Rightarrow \quad \mu = 0$$

From equation (1): $$\beta = 3(0) - 1 = -1$$ (valid since $$\beta < 0$$)

From equation (2): $$\alpha = \frac{2(0) - 39}{3} = \frac{-39}{3} = -13$$

Thus, $$\alpha = -13$$, $$\beta = -1$$, and:

$$\alpha^2 + \beta^2 = (-13)^2 + (-1)^2 = 169 + 1 = 170$$

Verification: Point P is $$(43, -13, -1)$$, point Q for $$\mu = 0$$ is $$(4, 0, -1)$$. Vector $$\overrightarrow{PQ} = (4 - 43, 0 - (-13), -1 - (-1)) = (-39, 13, 0)$$. Magnitude is $$\sqrt{(-39)^2 + 13^2} = \sqrt{1521 + 169} = \sqrt{1690} = 13\sqrt{10}$$, and direction ratios $$(-39, 13, 0)$$ are proportional to $$(3, -1, 0)$$.

The answer is 170.