Interpreting the Problem Statement

The question asks for the value of $\alpha^{2} + \beta^{2}$ for a point P(43, $\alpha$, $\beta$) where $\beta < 0$. We are given the distance from P to the line $\overrightarrow{\text{r}} = 4\hat{i} - \hat{k} + \mu(2\hat{i} + 3\hat{k})$ is $13\sqrt{10}$, measured *along* a line with direction ratios $3, -1, 0$. This means we need to find the intersection point (Q) of the given line (L) and a line passing through P with the direction $(3, -1, 0)$, and the distance PQ is $13\sqrt{10}$.

Line Representation

The given line (L) is represented by:

$\overrightarrow{\text{r}} = \overrightarrow{a} + \mu\overrightarrow{d}$

Where:

• $\overrightarrow{a} = 4\hat{i} - \hat{k} = (4, 0, -1)$ (a point on the line)
• $\overrightarrow{d} = 2\hat{i} + 3\hat{k} = (2, 0, 3)$ (direction vector of the line)

The point P is $(43, \alpha, \beta)$.

The direction ratios for the distance measurement are $\overrightarrow{v} = (3, -1, 0)$.

Finding the Intersection Point

Let the line passing through P with direction $\overrightarrow{v}$ be L'. Its equation is:

$\overrightarrow{\text{r}}' = P + t\overrightarrow{v} = (43, \alpha, \beta) + t(3, -1, 0)$

At the intersection point Q, the position vectors are equal:

$\overrightarrow{a} + \mu\overrightarrow{d} = P + t\overrightarrow{v}$

$(4, 0, -1) + \mu(2, 0, 3) = (43, \alpha, \beta) + t(3, -1, 0)$

Equating components:

1. $4 + 2\mu = 43 + 3t \implies 2\mu - 3t = 39$
2. $0 + 0\mu = \alpha - t \implies t = \alpha$
3. $-1 + 3\mu = \beta + 0t \implies 3\mu = \beta + 1$

Substitute $t = \alpha$ from (2) into (1):

$2\mu - 3\alpha = 39 \implies \mu = \frac{39 + 3\alpha}{2}$

From (3), we have:

$\mu = \frac{\beta + 1}{3}$

Equating the expressions for $\mu$:

$\frac{39 + 3\alpha}{2} = \frac{\beta + 1}{3}$

Cross-multiply:

$3(39 + 3\alpha) = 2(\beta + 1)$

$117 + 9\alpha = 2\beta + 2$

$9\alpha - 2\beta = -115$ (Equation 4)

Using the Distance Information

The distance between P and the intersection point Q is measured along the direction $\overrightarrow{v}$. This distance is given by $\| P - Q \| = \| t\overrightarrow{v} \| = |t| \|\overrightarrow{v}\|$.

Calculate the magnitude of $\overrightarrow{v}$:

$\|\overrightarrow{v}\| = \| (3, -1, 0) \| = \sqrt{3^2 + (-1)^2 + 0^2} = \sqrt{9 + 1} = \sqrt{10}$

We are given the distance is $13\sqrt{10}$:

$|t| \|\overrightarrow{v}\| = 13\sqrt{10}$

$|t| \sqrt{10} = 13\sqrt{10}$

$|t| = 13 \implies t = 13 \text{ or } t = -13$

Since $t = \alpha$, we have $\alpha = 13$ or $\alpha = -13$.

Solving for $\alpha$ and $\beta$

Use Equation 4 ($9\alpha - 2\beta = -115$) and the possible values of $\alpha$. We are also given the constraint $\beta < 0$.

• Case 1: $\alpha = 13$

  $9(13) - 2\beta = -115$

  $117 - 2\beta = -115$

  $2\beta = 117 + 115 = 232 \implies \beta = 116$

  This contradicts the condition $\beta < 0$.

• Case 2: $\alpha = -13$

  $9(-13) - 2\beta = -115$

  $-117 - 2\beta = -115$

  $-2\beta = -115 + 117 = 2 \implies \beta = -1$

  This satisfies the condition $\beta < 0$.

Therefore, the values are $\alpha = -13$ and $\beta = -1$.

Calculating $\alpha^{2} + \beta^{2}$

Finally, calculate $\alpha^{2} + \beta^{2}$:

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