If the point of intersection of the line
`vecr = (hati + 2 hatj + 3 hatk ) + lambda( 2 hati + hatj+ 2hatk )` and the plane `vecr (2 hati - 6 hatj + 3 hatk) + 5=0` lies on the plane `vec r ( hati + 75 hatj + 60 hatk) -alpha =0,` then `19 alpha + 17` is equal to :

Options

A. 1555
B. 1550
C. 2563
D. 2000

Answer

To solve the problem, we need to find the value of \(19\alpha + 17\) given the conditions of the line and the planes. Let's break down the solution step by step.

### Step 1: Identify the Line and Plane Equations
The line is given by:
\[ \vec{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(2\hat{i} + \hat{j} + 2\hat{k}) \]
This can be expressed in parametric form as:
\[ x = 1 + 2\lambda, \quad y = 2 + \lambda, \quad z = 3 + 2\lambda \]

The plane is given by:
\[ \vec{r} \cdot (2\hat{i} - 6\hat{j} + 3\hat{k}) + 5 = 0 \]
In Cartesian form, this translates to:
\[ 2x - 6y + 3z + 5 = 0 \]

### Step 2: Substitute Line Equations into the Plane Equation
We substitute the parametric equations of the line into the plane equation:
\[ 2(1 + 2\lambda) - 6(2 + \lambda) + 3(3 + 2\lambda) + 5 = 0 \]
Expanding this gives:
\[ 2 + 4\lambda - 12 - 6\lambda + 9 + 6\lambda + 5 = 0 \]
Combining like terms:
\[ 2 - 12 + 9 + 5 + (4\lambda - 6\lambda + 6\lambda) = 0 \]
This simplifies to:
\[ 4 - 6\lambda = 0 \]
Thus:
\[ -6\lambda + 4 = 0 \implies 6\lambda = 4 \implies \lambda = \frac{2}{3} \]

### Step 3: Find the Point of Intersection
Now we substitute \(\lambda = \frac{2}{3}\) back into the parametric equations to find the coordinates of the intersection point \(P\):
\[ x = 1 + 2\left(\frac{2}{3}\right) = 1 + \frac{4}{3} = \frac{7}{3} \]
\[ y = 2 + \frac{2}{3} = \frac{6}{3} + \frac{2}{3} = \frac{8}{3} \]
\[ z = 3 + 2\left(\frac{2}{3}\right) = 3 + \frac{4}{3} = \frac{9}{3} + \frac{4}{3} = \frac{13}{3} \]
So, the coordinates of point \(P\) are:
\[ P\left(\frac{7}{3}, \frac{8}{3}, \frac{13}{3}\right) \]

### Step 4: Verify Point Lies on the Second Plane
The second plane is given by:
\[ \vec{r} \cdot (\hat{i} + 75\hat{j} + 60\hat{k}) - \alpha = 0 \]
In Cartesian form, this is:
\[ x + 75y + 60z - \alpha = 0 \]
Substituting the coordinates of point \(P\):
\[ \frac{7}{3} + 75\left(\frac{8}{3}\right) + 60\left(\frac{13}{3}\right) - \alpha = 0 \]
Calculating:
\[ \frac{7}{3} + \frac{600}{3} + \frac{780}{3} - \alpha = 0 \]
Combining terms:
\[ \frac{7 + 600 + 780}{3} - \alpha = 0 \implies \frac{1387}{3} - \alpha = 0 \implies \alpha = \frac{1387}{3} \]

### Step 5: Calculate \(19\alpha + 17\)
Now we calculate \(19\alpha + 17\):
\[ 19\alpha + 17 = 19\left(\frac{1387}{3}\right) + 17 = \frac{26353}{3} + 17 = \frac{26353 + 51}{3} = \frac{26404}{3} \]

### Final Result
Thus, the value of \(19\alpha + 17\) is:
\[ \frac{26404}{3} \]