Step 1: Understanding the Question:
Three points on a line are collinear. The crossing point on the \( yz \)-plane has \( x = 0 \).
Step 2: Key Formula or Approach:
Use the condition that vectors formed by these points are parallel (direction ratios are proportional). Let \( P_1(a, 1, 6) \), \( P_2(3, 4, b) \), and \( P_3(0, 8.5, -6.5) \).
Step 3: Detailed Explanation:
Direction ratios of line \( P_2P_3 \):
\( (3-0, 4-8.5, b-(-6.5)) = (3, -4.5, b+6.5) \).
Direction ratios of line \( P_1P_2 \):
\( (a-3, 1-4, 6-b) = (a-3, -3, 6-b) \).
Since they are the same line, the DRs are proportional:
\[ \frac{a-3}{3} = \frac{-3}{-4.5} = \frac{6-b}{b+6.5} \]
From the middle term: \( \frac{-3}{-4.5} = \frac{3}{4.5} = \frac{30}{45} = \frac{2}{3} \).
Solving for \( a \):
\[ \frac{a-3}{3} = \frac{2}{3} \implies a-3 = 2 \implies a = 5 \]
Solving for \( b \):
\[ \frac{6-b}{b+6.5} = \frac{2}{3} \implies 3(6-b) = 2(b+6.5) \]
\[ 18 - 3b = 2b + 13 \implies 5b = 5 \implies b = 1 \]
Calculate \( 3a + 4b \):
\[ 3(5) + 4(1) = 15 + 4 = 19 \]
Step 4: Final Answer:
The value is 19.